| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the maximum x-velocity, we first need to find the velocity function from the given acceleration function. The acceleration function is given as a_x = 2.0 m/s^2 - (0.10 m / s^3) t We integrate this function with respect to time to obtain the velocity function: v_x = ∫ a_x dt = ∫ (2.0 m/s^2 - (0.10 m / s^3) t) dt v_x = 2.0t m/s - (0.10/2)t^2 m/s + C To find the constant of integration C, we use the initial condition that Sally's velocity is 0 m/s when t = 0 s. Thus, v_x(0) = 2.0(0) m/s - (0.10/2)(0)^2 m/s + C = 0 m/s C = 0 m/s So the velocity function is v_x = 2.0t m/s - (0.10/2)t^2 m/s To find the maximum velocity, we need to find when the derivative of v_x with respect to time is zero. This gives the time at which the velocity reaches its maximum: dv_x/dt = 2.0 m/s - (0.10 m / s^3) t = 0 Solving for t, we get t = 2.0 m/s / (0.10 m / s^3) = 20 s Now we can find the maximum velocity by plugging this time into the velocity function: v_x(max) = 2.0(20) m/s - (0.10/2)(20)^2 m/s = 40 m/s - 20 m/s = 20 m/s Finally, we can find the position at which this maximum velocity occurs by integrating the velocity function with respect to time: x = ∫ v_x dt = ∫ (2.0t m/s - (0.10/2)t^2 m/s) dt x = (2.0/2)t^2 m - (0.10/6)t^3 m + D To find the constant of integration D, we use the initial condition that Sally's position is 0 m when t = 0 s. Thus, x(0) = (2.0/2)(0)^2 m - (0.10/6)(0)^3 m + D = 0 m D = 0 m So the position function is x = (2.0/2)t^2 m - (0.10/6)t^3 m To find the position at which the maximum velocity occurs, we plug in the time t = 20 s: x(max) = (2.0/2)(20)^2 m - (0.10/6)(20)^3 m = 400 m - 133.3 m = 266.7 m Therefore, the car reaches its maximum x-velocity of 20 m/s at a position of X = 266.7 meters. |
517 |
| deepseekr1 | To determine the position \( X \) where Sally's car reaches its maximum velocity, we start by analyzing the given acceleration function and integrating it to find velocity and position as functions of time. **Given:** \[ a_x(t) = 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3) t \] **Step 1: Find the time \( t \) when acceleration becomes zero (maximum velocity occurs here).** \[ 0 = 2.0 - 0.10t \implies t = \frac{2.0}{0.10} = 20 \, \text{seconds} \] **Step 2: Integrate acceleration to find velocity \( v(t) \).** Assuming initial velocity \( v_0 = 0 \) (as the problem lacks this data but implies motion starts at the signpost): \[ v(t) = \int a_x(t) \, dt = 2.0t - 0.05t^2 + v_0 = 2.0t - 0.05t^2 \] **Step 3: Integrate velocity to find position \( x(t) \).** Assuming initial position \( x_0 = 0 \): \[ x(t) = \int v(t) \, dt = \int (2.0t - 0.05t^2) \, dt = 1.0t^2 - \frac{0.05}{3}t^3 \] **Step 4: Calculate \( X \) at \( t = 20 \, \text{s} \):** \[ X = 1.0(20)^2 - \frac{0.05}{3}(20)^3 = 400 - \frac{0.05 \times 8000}{3} = 400 - \frac{400}{3} = \frac{800}{3} \approx 266.67 \, \text{m} \] **Answer:** \[ \boxed{267 \, \text{meters}} \] |
517 |
| gpto1 | To solve this problem, we need to determine the position \( X \) at which Sally's car reaches its maximum velocity along the x-direction. The acceleration is given as a function of time: \[ a_x(t) = 2.0\, \text{m/s}^2 - 0.10\, \text{m/s}^3 \times t \] **Step 1: Find the Time when Velocity is Maximum** The car's velocity will be at a maximum when the acceleration changes from positive to negative, i.e., when \( a_x(t) = 0 \). Set the acceleration equal to zero and solve for \( t \): \[ \begin{align*} a_x(t) &= 0 \\ 2.0\, \text{m/s}^2 - 0.10\, \text{m/s}^3 \times t &= 0 \\ 0.10\, \text{m/s}^3 \times t &= 2.0\, \text{m/s}^2 \\ t &= \frac{2.0\, \text{m/s}^2}{0.10\, \text{m/s}^3} \\ t &= 20\, \text{s} \end{align*} \] So, the car reaches its maximum velocity at \( t = 20\, \text{s} \). **Step 2: Find the Velocity as a Function of Time** Integrate the acceleration to get the velocity. Assume the initial velocity \( v(0) = 0\, \text{m/s} \) (since the initial velocity is not specified): \[ \begin{align*} v(t) &= \int a_x(t) \, dt \\ &= \int (2.0 - 0.10 t) \, dt \\ &= 2.0 t - 0.05 t^2 + C_v \end{align*} \] Using the initial condition \( v(0) = 0 \): \[ 0 = 2.0 \times 0 - 0.05 \times 0^2 + C_v \implies C_v = 0 \] So, the velocity function is: \[ v(t) = 2.0 t - 0.05 t^2 \] **Step 3: Find the Position as a Function of Time** Integrate the velocity to get the position. Again, assume the initial position \( x(0) = 0 \): \[ \begin{align*} x(t) &= \int v(t) \, dt \\ &= \int (2.0 t - 0.05 t^2) \, dt \\ &= t^2 - \frac{0.05}{3} t^3 + C_x \end{align*} \] Using the initial condition \( x(0) = 0 \): \[ 0 = 0 - 0 + C_x \implies C_x = 0 \] Simplify the position function: \[ x(t) = t^2 - \frac{t^3}{60} \] **Step 4: Calculate the Position at \( t = 20\, \text{s} \)** Plug \( t = 20\, \text{s} \) into the position function: \[ \begin{align*} x(20) &= (20)^2 - \frac{(20)^3}{60} \\ &= 400 - \frac{8000}{60} \\ &= 400 - \frac{8000}{60} \\ &= 400 - \frac{400}{3} \\ &= \frac{1200}{3} - \frac{400}{3} \\ &= \frac{800}{3} \\ &\approx 266.67\, \text{m} \end{align*} \] **Answer:** At position \( X = \dfrac{800}{3} \) meters (approximately 266.67 m), the car reaches its maximum x-velocity. |
517 |
| GPT-4o | To determine the distance \( X \) at which Sally's 1965 Mustang reaches its maximum x-velocity, we need to follow these steps: 1. **Find the time \( t \) at which the x-velocity \( v_x \) is maximized**: The x-acceleration \( a_x \) is given by: \[ a_x = 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3) \cdot t \] The x-velocity \( v_x \) reaches a maximum when the acceleration \( a_x \) is zero. Set \( a_x \) to zero and solve for \( t \): \[ 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3) \cdot t = 0 \] \[ (0.10 \, \text{m/s}^3) \cdot t = 2.0 \, \text{m/s}^2 \] \[ t = \frac{2.0 \, \text{m/s}^2}{0.10 \, \text{m/s}^3} = 20 \, \text{s} \] 2. **Determine the x-velocity \( v_x \) as a function of time**: Integrate the acceleration function \( a_x \) with respect to time to get the velocity function \( v_x \): \[ v_x(t) = \int a_x \, dt = \int \left(2.0 \, \text{m/s}^2 - 0.10 \, \text{m/s}^3 \cdot t\right) dt \] \[ v_x(t) = 2.0 \, t - 0.05 \, t^2 + C \] Given that Sally passes the signpost at \( t = 0 \) with an initial velocity \( v_{x0} \), we can find the constant \( C \). For simplicity, let's assume \( v_{x0} = 0 \) (if not otherwise specified): \[ v_x(0) = 0 \implies C = 0 \] \[ v_x(t) = 2.0 \, t - 0.05 \, t^2 \] 3. **Evaluate the x-velocity at \( t = 20 \) seconds**: \[ v_x(20) = 2.0 \cdot 20 - 0.05 \cdot (20)^2 \] \[ v_x(20) = 40 - 0.05 \cdot 400 \] \[ v_x(20) = 40 - 20 = 20 \, \text{m/s} \] 4. **Determine the x-position \( X \) at time \( t = 20 \) seconds**: Integrate the velocity function \( v_x(t) \) to get the position function \( x(t) \): \[ x(t) = \int v_x(t) \, dt = \int \left(2.0 \, t - 0.05 \, t^2\right) dt \] \[ x(t) = 2.0 \frac{t^2}{2} - 0.05 \frac{t^3}{3} + D \] \[ x(t) = t^2 - \frac{0.05}{3} t^3 + D \] Given that Sally passes the signpost at \( t = 0 \) with an initial position \( x_0 \), we can find the constant \( D \). Assume \( x_0 = 0 \) (if not otherwise specified): \[ x(0) = 0 \implies D = 0 \] \[ x(t) = t^2 - \frac{0.05}{3} t^3 \] Evaluate \( x(t) \) at \( t = 20 \) seconds: \[ x(20) = (20)^2 - \frac{0.05}{3} \cdot (20)^3 \] \[ x(20) = 400 - \frac{0.05}{3} \cdot 8000 \] \[ x(20) = 400 - \frac{400}{3} \] \ |
517 |
| llama3.1-70B-Instruct | To find the position X at which the car reaches its maximum x-velocity, we need to find the time at which the velocity is maximum, and then use that time to find the position. First, we need to find the velocity as a function of time. We can do this by integrating the acceleration with respect to time: v_x(t) = ∫a_x(t) dt = ∫(2.0 m/s^2 - (0.10 m/s^3)t) dt = 2.0t - (0.05 m/s^2)t^2 + C where C is the constant of integration. We know that the initial velocity is 20 m/s, so we can set t = 0 and v_x = 20 m/s to find C: 20 m/s = 2.0(0) - (0.05 m/s^2)(0)^2 + C C = 20 m/s So, the velocity as a function of time is: v_x(t) = 2.0t - (0.05 m/s^2)t^2 + 20 m/s To find the time at which the velocity is maximum, we can take the derivative of the velocity with respect to time and set it equal to zero: dv_x/dt = 2.0 - (0.10 m/s^2)t = 0 Solving for t, we get: t = 20 s Now, we can use this time to find the position X at which the car reaches its maximum x-velocity. We can do this by integrating the velocity with respect to time: X = ∫v_x(t) dt = ∫(2.0t - (0.05 m/s^2)t^2 + 20 m/s) dt = (1.0 m/s)t^2 - (0.0167 m/s^2)t^3 + 20t + D where D is the constant of integration. We know that the initial position is 0 m, so we can set t = 0 and X = 0 to find D: 0 m = (1.0 m/s)(0)^2 - (0.0167 m/s^2)(0)^3 + 20(0) + D D = 0 m So, the position as a function of time is: X = (1.0 m/s)t^2 - (0.0167 m/s^2)t^3 + 20t Now, we can plug in the time t = 20 s to find the position X: X = (1.0 m/s)(20 s)^2 - (0.0167 m/s^2)(20 s)^3 + 20(20 s) = 400 m - 133.3 m + 400 m = 666.7 m Therefore, the car reaches its maximum x-velocity at a position X = 666.7 m. |
517 |
| llama3.1-405B-Instruct | To find the maximum x-velocity, we need to find the time at which the acceleration is zero, since the acceleration is the rate of change of velocity. We can set the acceleration equation equal to zero and solve for time: a_x = 2.0 m/s^2 - (0.10 m / s^3) t = 0 Solving for t, we get: t = 2.0 m/s^2 / (0.10 m / s^3) = 20 s Now, we need to find the distance traveled by Sally's car from the signpost to the point where the car reaches maximum x-velocity. We can use the equation of motion: x = x0 + v0t + (1/2)at^2 where x0 = 0 (since we're measuring from the signpost), v0 = 25 m/s (the initial velocity), and a = 2.0 m/s^2 (the acceleration). Plugging in the values, we get: x = 0 + (25 m/s)(20 s) + (1/2)(2.0 m/s^2)(20 s)^2 x = 500 m + 400 m x = 900 m Therefore, the car reaches maximum x-velocity at X = 900 meters. |
517 |
| DeepSeek-Coder-V2 | To solve the problem of finding the position \( X \) where Sally's car reaches its maximum x-velocity, we need to follow these steps: 1. **Determine the time \( t \) when the acceleration \( a_x \) is zero.** 2. **Find the maximum velocity \( v_x \) at that time.** 3. **Calculate the position \( X \) at the time when the velocity is maximum.** ### Step 1: Find the time \( t \) when \( a_x = 0 \) The acceleration \( a_x \) is given by: \[ a_x = 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3) t \] Set \( a_x = 0 \): \[ 2.0 - 0.10t = 0 \] Solve for \( t \): \[ 0.10t = 2.0 \] \[ t = \frac{2.0}{0.10} \] \[ t = 20 \, \text{s} \] ### Step 2: Find the maximum velocity \( v_x \) at \( t = 20 \, \text{s} \) The velocity \( v_x \) is the integral of the acceleration \( a_x \) with respect to time: \[ v_x = \int a_x \, dt \] \[ v_x = \int (2.0 - 0.10t) \, dt \] \[ v_x = 2.0t - 0.05t^2 + C \] Assuming the initial velocity \( v_0 = 0 \) at \( t = 0 \): \[ v_x = 2.0t - 0.05t^2 \] Evaluate \( v_x \) at \( t = 20 \, \text{s} \): \[ v_x = 2.0(20) - 0.05(20)^2 \] \[ v_x = 40 - 0.05(400) \] \[ v_x = 40 - 20 \] \[ v_x = 20 \, \text{m/s} \] ### Step 3: Calculate the position \( X \) at \( t = 20 \, \text{s} \) The position \( x \) is the integral of the velocity \( v_x \) with respect to time: \[ x = \int v_x \, dt \] \[ x = \int (2.0t - 0.05t^2) \, dt \] \[ x = 1.0t^2 - 0.0167t^3 + C \] Assuming the initial position \( x_0 = 0 \) at \( t = 0 \): \[ x = 1.0t^2 - 0.0167t^3 \] Evaluate \( x \) at \( t = 20 \, \text{s} \): \[ x = 1.0(20)^2 - 0.0167(20)^3 \] \[ x = 1.0(400) - 0.0167(8000) \] \[ x = 400 - 133.6 \] \[ x = 266.4 \, \text{m} \] Thus, the position \( X \) where the car reaches its maximum x-velocity is: \[ \boxed{266.4} \] |
517 |
| Qwen2-72B-Instruct | To find the maximum x-velocity, we first need to find the velocity function from the given acceleration function. The acceleration function is given as a_x = 2.0 m/s^2 - (0.10 m / s^3) t We integrate this function with respect to time to obtain the velocity function: v_x = ∫ a_x dt = ∫ (2.0 m/s^2 - (0.10 m / s^3) t) dt v_x = 2.0t m/s - (0.10/2)t^2 m/s + C To find the constant of integration C, we use the initial condition that Sally's velocity is 0 m/s when t = 0 s. Thus, v_x(0) = 2.0(0) m/s - (0.10/2)(0)^2 m/s + C = 0 m/s C = 0 m/s So the velocity function is v_x = 2.0t m/s - (0.10/2)t^2 m/s To find the maximum velocity, we need to find when the derivative of v_x with respect to time is zero. This gives the time at which the velocity reaches its maximum: dv_x/dt = 2.0 m/s - (0.10 m / s^3) t = 0 Solving for t, we get t = 2.0 m/s / (0.10 m / s^3) = 20 s Now we can find the maximum velocity by plugging this time into the velocity function: v_x(max) = 2.0(20) m/s - (0.10/2)(20)^2 m/s = 40 m/s - 20 m/s = 20 m/s Finally, we can find the position at which this maximum velocity occurs by integrating the velocity function with respect to time: x = ∫ v_x dt = ∫ (2.0t m/s - (0.10/2)t^2 m/s) dt x = (2.0/2)t^2 m - (0.10/6)t^3 m + D To find the constant of integration D, we use the initial condition that Sally's position is 0 m when t = 0 s. Thus, x(0) = (2.0/2)(0)^2 m - (0.10/6)(0)^3 m + D = 0 m D = 0 m So the position function is x = (2.0/2)t^2 m - (0.10/6)t^3 m To find the position at which the maximum velocity occurs, we plug in the time t = 20 s: x(max) = (2.0/2)(20)^2 m - (0.10/6)(20)^3 m = 400 m - 133.3 m = 266.7 m Therefore, the car reaches its maximum x-velocity of 20 m/s at a position of X = 266.7 meters. |
517 |
| baichuan4 | To solve for the position (X) where Sally's car reaches maximum x-velocity, we need to consider the acceleration equation provided and find the time (t) at which the acceleration becomes zero since the velocity will be at a maximum when the acceleration is zero. The acceleration equation is: \[a_x = 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3)t\] We set \(a_x = 0\) to find the time at which the car stops accelerating: \[0 = 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3)t\] Solving for t: \[(0.10 \, \text{m/s}^3)t = 2.0 \, \text{m/s}^2\] \[t = \frac{2.0 \, \text{m/s}^2}{0.10 \, \text{m/s}^3}\] \[t = 20 \, \text{s}\] Now that we know the time at which the car reaches maximum velocity, we can calculate the position (X) by using the kinematic equation: \[v_f = v_i + at\] We are given an initial velocity (\(v_i\)) of 15 m/s in the positive x-direction. We found the time (t) to be 20 seconds, and the acceleration (a) before it becomes zero is \(2.0 \, \text{m/s}^2\). Plugging in the values: \[v_f = 15 \, \text{m/s} + (2.0 \, \text{m/s}^2)(20 \, \text{s})\] \[v_f = 15 \, \text{m/s} + 40 \, \text{m/s}\] \[v_f = 55 \, \text{m/s}\] Now, to find the position (X), we use the equation: \[X = X_0 + v_i t + \frac{1}{2}at^2\] Given \(X_0 = 0\) since we start at the signpost, \(v_i = 15 \, \text{m/s}\), \(a = 2.0 \, \text{m/s}^2\), and \(t = 20 \, \text{s}\): \[X = 0 + (15 \, \text{m/s})(20 \, \text{s}) + \frac{1}{2}(2.0 \, \text{m/s}^2)(20 \, \text{s})^2\] \[X = 300 \, \text{m} + 400 \, \text{m}\] \[X = 700 \, \text{m}\] So, the car reaches its maximum x-velocity at 700 meters from the signpost. |
517 |
| doubao | First find the time when the velocity is maximum. The derivative of velocity is acceleration. When acceleration is zero, velocity is at its maximum. Set a_x = 0: 0 = 2.0 - (0.10)t 0.10t = 2.0 t = 20 s. To find the velocity at this time, integrate the acceleration function: v_x = ∫(2.0 - (0.10t))dt = 2.0t - 0.05t^2 + C (where C is the initial velocity which is not given, but we don't need it for this part). Evaluating at t = 20 s gives: v_x = 2.0*20 - 0.05*20^2 = 20 m/s. Now to find the position X where the car reaches this maximum velocity, integrate the velocity function: x = ∫v_x dt = ∫(2.0t - 0.05t^2)dt = t^2 - 0.0167t^3 + D (where D is the initial position which is not given). Evaluating at t = 20 s gives: X = 20^2 - 0.0167*20^3 = 200 m. |
517 |
| minimax | To find the position \( X \) at which Sally's car reaches its maximum x-velocity, we need to understand that the maximum velocity occurs at the point where the acceleration becomes zero, because the car will start decelerating after reaching its peak velocity. Given the acceleration function: \[ a_x = 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3) \cdot t \] We set the acceleration to zero to find the time \( t \) at which the maximum velocity occurs: \[ 0 = 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3) \cdot t \] Solving for \( t \): \[ (0.10 \, \text{m/s}^3) \cdot t = 2.0 \, \text{m/s}^2 \] \[ t = \frac{2.0 \, \text{m/s}^2}{0.10 \, \text{m/s}^3} \] \[ t = 20 \, \text{s} \] Now that we have the time at which the maximum velocity occurs, we can find the position \( X \) by integrating the velocity function over time. First, we need to find the velocity function by integrating the acceleration function: \[ v_x(t) = \int a_x(t) \, dt \] \[ v_x(t) = \int (2.0 \, \text{m/s}^2 - 0.10 \, \text{m/s}^3 \cdot t) \, dt \] \[ v_x(t) = (2.0 \, \text{m/s}^2)t - \frac{0.10 \, \text{m/s}^3}{2} \cdot t^2 + C \] Since the car starts from rest, the initial velocity \( v_x(0) \) is zero, which means the constant of integration \( C \) is zero. So the velocity function simplifies to: \[ v_x(t) = (2.0 \, \text{m/s}^2)t - \frac{0.10 \, \text{m/s}^3}{2} \cdot t^2 \] Now we plug in the time \( t = 20 \, \text{s} \) to find the maximum velocity: \[ v_x(20 \, \text{s}) = (2.0 \, \text{m/s}^2)(20 \, \text{s}) - \frac{0.10 \, \text{m/s}^3}{2} \cdot (20 \, \text{s})^2 \] \[ v_x(20 \, \text{s}) = 40 \, \text{m/s} - \frac{0.10 \, \text{m/s}^3}{2} \cdot 400 \, \text{s}^2 \] \[ v_x(20 \, \text{s}) = 40 \, \text{m/s} - 20 \, \text{m/s} \] \[ v_x(20 \, \text{s}) = 20 \, \text{m/s} \] Finally, we integrate the velocity function to find the position \( X \): \[ X(t) = \int v_x(t) \, dt \] \[ X(t) = \int [(2.0 \, \text{m/s}^2)t - \frac{0.10 \, \text{m/s}^3}{2} \cdot t^2] \, dt \] \[ X(t) = \frac{2.0 \, \text{m/s}^2}{2} \cdot t^2 - \frac{0.10 \, \text{m/s}^3}{6} \cdot t^3 + C' \] Again, since the car starts from the origin, the initial position \( X(0) \) is zero, which means the constant of integration \( C' \) is zero. So the position function simplifies to: \[ X(t) = \frac{2.0 \, \text{m/s}^2}{2} \cdot t^2 - \frac{0.10 \, \text{m/s}^3}{6} \cdot t^3 \] \[ X(t) = 1.0 \, \text{m/s}^2 \cdot t^2 - \frac{0.10 \, \text{m/s}^3}{6} \cdot t^3 \] Now we plug in the time \( t = 20 \, \text{s} \) to find the position \( X \): \[ X(20 \, \text{s}) = 1.0 \, \text{m/s}^2 \cdot (20 \, \text{s})^2 - \frac{0.10 \, \text{m/s}^3}{6} \cdot (20 \, \text{s})^3 \] \[ X(20 \, \text{s}) = 1.0 \, \text{m/s}^2 \cdot 400 \, \text{s}^2 - \frac{0.10 \, \text{m/s}^3}{6} \cdot 8000 \, \text{s}^3 \] \[ X(20 \, \text{s}) = 400 \, \text{m} - \frac{800}{6} \, \text{m} \] \[ X(20 \, \text{s}) = 400 \, \text{m} - 133.\overline{3} \, \text{m} \] \[ X(20 \, \text{s}) = 266.\overline{6} \, \text{m} \] Therefore, Sally's car reaches its maximum x-velocity at a position of approximately 266.7 meters along the highway. |
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| kimi | To find the maximum x-velocity and the corresponding x-position, we first need to find the velocity function \( v_x(t) \) by integrating the acceleration function \( a_x(t) \) with respect to time \( t \). Given: \[ a_x(t) = 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3) t \] Integrating \( a_x(t) \) with respect to \( t \) to get \( v_x(t) \): \[ v_x(t) = \int a_x(t) \, dt \] \[ v_x(t) = \int (2.0 \, \text{m/s}^2 - 0.10 \, \text{m/s}^3 t) \, dt \] \[ v_x(t) = 2.0t - 0.10 \frac{t^2}{2} + C_1 \] \[ v_x(t) = 2.0t - 0.05t^2 + C_1 \] To find \( C_1 \), we need an initial condition. Let's assume Sally starts from rest at \( t = 0 \), so \( v_x(0) = 0 \): \[ 0 = 2.0(0) - 0.05(0)^2 + C_1 \] \[ C_1 = 0 \] So the velocity function is: \[ v_x(t) = 2.0t - 0.05t^2 \] The maximum velocity occurs when the derivative of the velocity function with respect to time is zero, i.e., when \( a_x(t) = 0 \): \[ 0 = 2.0 - 0.10t \] \[ t = \frac{2.0}{0.10} \] \[ t = 20 \, \text{s} \] Now we can find the maximum velocity by plugging \( t = 20 \) into the velocity function: \[ v_{x_{max}} = 2.0(20) - 0.05(20)^2 \] \[ v_{x_{max}} = 40 - 0.05(400) \] \[ v_{x_{max}} = 40 - 20 \] \[ v_{x_{max}} = 20 \, \text{m/s} \] To find the position \( X \) at which this maximum velocity occurs, we integrate the velocity function with respect to time: \[ x(t) = \int v_x(t) \, dt \] \[ x(t) = \int (2.0t - 0.05t^2) \, dt \] \[ x(t) = t^2 - 0.05 \frac{t^3}{3} + C_2 \] To find \( C_2 \), we need another initial condition. Let's assume Sally starts at the origin \( x = 0 \) at \( t = 0 \): \[ 0 = (0)^2 - 0.05 \frac{(0)^3}{3} + C_2 \] \[ C_2 = 0 \] So the position function is: \[ x(t) = t^2 - 0.05 \frac{t^3}{3} \] Now we can find the position \( X \) at \( t = 20 \) seconds: \[ X = (20)^2 - 0.05 \frac{(20)^3}{3} \] \[ X = 400 - 0.05 \frac{8000}{3} \] \[ X = 400 - \frac{400}{3} \] \[ X = 400 - 133.33 \] \[ X = 266.67 \, \text{m} \] So, Sally reaches a maximum x-velocity of \( 20 \, \text{m/s} \) at \( X = 266.67 \, \text{m} \). |
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| wenxinyiyan | To find the time at which the car reaches its maximum velocity, we need to find the time when the acceleration is zero. So we set the acceleration function equal to zero and solve for t: 2.0 m/s^2 - (0.10 m/s^3) t = 0 Solving for t, we get: t = (2.0 m/s^2) / (0.10 m/s^3) = 20 s Now that we have the time at which the car reaches its maximum velocity, we can find the velocity at that time. The velocity as a function of time is given by the integral of the acceleration function with respect to time: v_x(t) = ∫ a_x(t) dt = ∫ (2.0 m/s^2 - (0.10 m/s^3) t) dt Integrating, we get: v_x(t) = 2.0 m/s^2 * t - (0.10 m/s^3) * t^2 / 2 + C Where C is the constant of integration. Since we know that the initial velocity is zero (the car starts from rest), we can set v_x(0) = 0 and solve for C: 0 = 2.0 m/s^2 * 0 - (0.10 m/s^3) * 0^2 / 2 + C C = 0 So the velocity function is: v_x(t) = 2.0 m/s^2 * t - (0.10 m/s^3) * t^2 / 2 Now we can find the maximum velocity by plugging in t = 20 s: v_x(20 s) = 2.0 m/s^2 * 20 s - (0.10 m/s^3) * (20 s)^2 / 2 v_x(20 s) = 40 m/s - 20 m/s v_x(20 s) = 20 m/s Finally, to find the distance X at which the car reaches its maximum velocity, we can use the equation of motion: X = v_0 * t + 1/2 * a * t^2 But since the acceleration is not constant, we need to use the average acceleration over the interval from 0 to 20 s. The average acceleration is given by: a_avg = (a_0 + a_f) / 2 Where a_0 is the initial acceleration and a_f is the final acceleration (which is zero at t = 20 s). So: a_avg = (2.0 m/s^2 + 0) / 2 = 1.0 m/s^2 Now we can find X: X = v_0 * t + 1/2 * a_avg * t^2 X = 0 * 20 s + 1/2 * 1.0 m/s^2 * (20 s)^2 X = 1/2 * 1.0 m/s^2 * 400 s^2 X = 200 m So the car reaches its maximum velocity of 20 m/s at a distance of 200 meters from the signpost. |
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| zhipu | To find the position \( X \) where Sally's car reaches its maximum x-velocity, we need to follow these steps: 1. **Determine the time \( t \) at which the maximum velocity occurs.** 2. **Calculate the position \( X \) at that time.** ### Step 1: Find the time \( t \) at which the maximum velocity occurs The velocity \( v_x \) is the integral of the acceleration \( a_x \) with respect to time \( t \): \[ a_x = \frac{dv_x}{dt} = 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3) t \] To find the velocity \( v_x \), we integrate \( a_x \): \[ v_x = \int a_x \, dt = \int (2.0 - 0.10 t) \, dt \] \[ v_x = 2.0 t - \frac{0.10 t^2}{2} + C \] Given that \( v_x = 10 \, \text{m/s} \) at \( t = 0 \): \[ 10 = 2.0 \cdot 0 - \frac{0.10 \cdot 0^2}{2} + C \] \[ C = 10 \] So, the velocity function is: \[ v_x = 2.0 t - 0.05 t^2 + 10 \] To find the maximum velocity, we take the derivative of \( v_x \) with respect to \( t \) and set it to zero: \[ \frac{dv_x}{dt} = 2.0 - 0.10 t = 0 \] Solving for \( t \): \[ 0.10 t = 2.0 \] \[ t = \frac{2.0}{0.10} = 20 \, \text{s} \] ### Step 2: Calculate the position \( X \) at \( t = 20 \, \text{s} \) The position \( x \) is the integral of the velocity \( v_x \) with respect to time \( t \): \[ v_x = \frac{dx}{dt} = 2.0 t - 0.05 t^2 + 10 \] Integrate \( v_x \) to find \( x \): \[ x = \int v_x \, dt = \int (2.0 t - 0.05 t^2 + 10) \, dt \] \[ x = \int 2.0 t \, dt - \int 0.05 t^2 \, dt + \int 10 \, dt \] \[ x = 2.0 \frac{t^2}{2} - 0.05 \frac{t^3}{3} + 10 t + D \] \[ x = t^2 - \frac{0.05 t^3}{3} + 10 t + D \] Given that \( x = 0 \) at \( t = 0 \): \[ 0 = 0^2 - \frac{0.05 \cdot 0^3}{3} + 10 \cdot 0 + D \] \[ D = 0 \] So, the position function is: \[ x = t^2 - \frac{0.05 t^3}{3} + 10 t \] Now, substitute \( t = 20 \, \text{s} \) into the position function: \[ x = (20)^2 - \frac{0.05 (20)^3}{3} + 10 (20) \] \[ x = 400 - \frac{0.05 \cdot 8000}{3} + 200 \] \[ x = 400 - \frac{400}{3} + 200 \] \[ x = 400 + 200 - \frac{400}{3} \] \[ x = 600 - \frac{400}{3} \] \[ x = 600 - 133.33 \] \[ x \approx 466.67 \, \text{m} \] Therefore, the position \( X \) where the car reaches its maximum x-velocity is approximately \( 466.67 \, \text{m} \). |
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| gpto1 | To solve this problem, we need to determine the position \( X \) at which Sally's car reaches its maximum velocity along the x-direction. The acceleration is given as a function of time: \[ a_x(t) = 2.0\, \text{m/s}^2 - 0.10\, \text{m/s}^3 \times t \] **Step 1: Find the Time when Velocity is Maximum** The car's velocity will be at a maximum when the acceleration changes from positive to negative, i.e., when \( a_x(t) = 0 \). Set the acceleration equal to zero and solve for \( t \): \[ \begin{align*} a_x(t) &= 0 \\ 2.0\, \text{m/s}^2 - 0.10\, \text{m/s}^3 \times t &= 0 \\ 0.10\, \text{m/s}^3 \times t &= 2.0\, \text{m/s}^2 \\ t &= \frac{2.0\, \text{m/s}^2}{0.10\, \text{m/s}^3} \\ t &= 20\, \text{s} \end{align*} \] So, the car reaches its maximum velocity at \( t = 20\, \text{s} \). **Step 2: Find the Velocity as a Function of Time** Integrate the acceleration to get the velocity. Assume the initial velocity \( v(0) = 0\, \text{m/s} \) (since the initial velocity is not specified): \[ \begin{align*} v(t) &= \int a_x(t) \, dt \\ &= \int (2.0 - 0.10 t) \, dt \\ &= 2.0 t - 0.05 t^2 + C_v \end{align*} \] Using the initial condition \( v(0) = 0 \): \[ 0 = 2.0 \times 0 - 0.05 \times 0^2 + C_v \implies C_v = 0 \] So, the velocity function is: \[ v(t) = 2.0 t - 0.05 t^2 \] **Step 3: Find the Position as a Function of Time** Integrate the velocity to get the position. Again, assume the initial position \( x(0) = 0 \): \[ \begin{align*} x(t) &= \int v(t) \, dt \\ &= \int (2.0 t - 0.05 t^2) \, dt \\ &= t^2 - \frac{0.05}{3} t^3 + C_x \end{align*} \] Using the initial condition \( x(0) = 0 \): \[ 0 = 0 - 0 + C_x \implies C_x = 0 \] Simplify the position function: \[ x(t) = t^2 - \frac{t^3}{60} \] **Step 4: Calculate the Position at \( t = 20\, \text{s} \)** Plug \( t = 20\, \text{s} \) into the position function: \[ \begin{align*} x(20) &= (20)^2 - \frac{(20)^3}{60} \\ &= 400 - \frac{8000}{60} \\ &= 400 - \frac{8000}{60} \\ &= 400 - \frac{400}{3} \\ &= \frac{1200}{3} - \frac{400}{3} \\ &= \frac{800}{3} \\ &\approx 266.67\, \text{m} \end{align*} \] **Answer:** At position \( X = \dfrac{800}{3} \) meters (approximately 266.67 m), the car reaches its maximum x-velocity. |
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| deepseekv3 | To solve this problem, we need to determine the distance \( X \) at which Sally's car reaches its maximum velocity. Here's a step-by-step solution: --- ### 1. **Given Acceleration Function** The acceleration as a function of time is: \[ a_x(t) = 2.0 \, \text{m/s}^2 - (0.10 \, \text{m/s}^3) t \] --- ### 2. **Find the Time When Acceleration is Zero** The car reaches its maximum velocity when the acceleration becomes zero. Set \( a_x(t) = 0 \) and solve for \( t \): \[ 2.0 - 0.10 t = 0 \] \[ t = \frac{2.0}{0.10} = 20 \, \text{s} \] --- ### 3. **Find the Velocity as a Function of Time** Integrate the acceleration function to find the velocity as a function of time: \[ v_x(t) = \int a_x(t) \, dt = \int (2.0 - 0.10 t) \, dt \] \[ v_x(t) = 2.0 t - 0.10 \frac{t^2}{2} + C \] Assuming the initial velocity is \( v_x(0) = 0 \), the constant \( C = 0 \): \[ v_x(t) = 2.0 t - 0.05 t^2 \] --- ### 4. **Find the Maximum Velocity** Substitute \( t = 20 \, \text{s} \) into the velocity function to find the maximum velocity: \[ v_{\text{max}} = 2.0 (20) - 0.05 (20)^2 \] \[ v_{\text{max}} = 40 - 20 = 20 \, \text{m/s} \] --- ### 5. **Find the Distance \( X \) as a Function of Time** Integrate the velocity function to find the distance as a function of time: \[ x(t) = \int v_x(t) \, dt = \int (2.0 t - 0.05 t^2) \, dt \] \[ x(t) = 1.0 t^2 - \frac{0.05}{3} t^3 + C \] Assuming the initial position is \( x(0) = 0 \), the constant \( C = 0 \): \[ x(t) = 1.0 t^2 - \frac{0.05}{3} t^3 \] --- ### 6. **Calculate \( X \) at \( t = 20 \, \text{s} \)** Substitute \( t = 20 \, \text{s} \) into the distance function: \[ X = 1.0 (20)^2 - \frac{0.05}{3} (20)^3 \] \[ X = 400 - \frac{0.05}{3} (8000) \] \[ X = 400 - 133.33 = 266.67 \, \text{m} \] --- ### Final Answer: The car reaches its maximum velocity at \( X = 266.67 \, \text{m} \). |
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