Question 1204877: The acceleration of a bus is given by ax(t) = at, where a = 1.2m/s^3.
(a) If the bus's velocity at time t = 1.0s is 5.0 m/s, what is its velocity at time t = 20 s?
(b) If the bus's position at time t = 1.0 s is 6.0 m, what is its position at time t = 2.0 s?
(c) Sketch ay-t, vy-t and x-t graphs for the motion.
problem from Young and Freedman. University Physics with Modern Physics. Fifteenth Edition.
Answer by asinus(45)   (Show Source):
You can put this solution on YOUR website! To solve the problem, we will analyze the acceleration function, integrate it to find the velocity function, and then integrate the velocity function to find the position function.
### Given Information
- Acceleration:
$$
a_x(t) = at = 1.2t \quad \text{(where } a = 1.2 \, \text{m/s}^3\text{)}
$$
### Part (a): Find the Velocity at $ t = 20 \, \text{s} $
1. **Integrate the acceleration to find the velocity**:
$$
v_x(t) = \int a_x(t) \, dt = \int 1.2t \, dt = 1.2 \cdot \frac{t^2}{2} + C = 0.6t^2 + C
$$
2. **Use the initial condition**:
Given that $ v_x(1) = 5.0 \, \text{m/s} $:
$$
5.0 = 0.6(1)^2 + C \implies 5.0 = 0.6 + C \implies C = 5.0 - 0.6 = 4.4
$$
3. **Velocity function**:
$$
v_x(t) = 0.6t^2 + 4.4
$$
4. **Calculate the velocity at $ t = 20 \, \text{s} $**:
$$
v_x(20) = 0.6(20)^2 + 4.4 = 0.6 \cdot 400 + 4.4 = 240 + 4.4 = 244.4 \, \text{m/s}
$$
### Part (b): Find the Position at $ t = 2.0 \, \text{s} $
1. **Integrate the velocity to find the position**:
$$
x(t) = \int v_x(t) \, dt = \int (0.6t^2 + 4.4) \, dt = 0.6 \cdot \frac{t^3}{3} + 4.4t + D = 0.2t^3 + 4.4t + D
$$
2. **Use the initial condition**:
Given that $ x(1) = 6.0 \, \text{m} $:
$$
6.0 = 0.2(1)^3 + 4.4(1) + D \implies 6.0 = 0.2 + 4.4 + D \implies D = 6.0 - 4.6 = 1.4
$$
3. **Position function**:
$$
x(t) = 0.2t^3 + 4.4t + 1.4
$$
4. **Calculate the position at $ t = 2.0 \, \text{s} $**:
$$
x(2) = 0.2(2)^3 + 4.4(2) + 1.4 = 0.2 \cdot 8 + 8.8 + 1.4 = 1.6 + 8.8 + 1.4 = 11.8 \, \text{m}
$$
### Part (c): Sketch the Graphs
1. **Acceleration vs. Time $ a_y(t) $**:
- The acceleration function is linear:
$$
a_x(t) = 1.2t
$$
- At $ t = 0 $, $ a_x(0) = 0 $ and at $ t = 20 $, $ a_x(20) = 24 \, \text{m/s}^2 $.
2. **Velocity vs. Time $ v_y(t) $**:
- The velocity function is quadratic:
$$
v_x(t) = 0.6t^2 + 4.4
$$
- At $ t = 1 $, $ v_x(1) = 5.0 \, \text{m/s} $ and at $ t = 20 $, $ v_x(20) = 244.4 \, \text{m/s} $.
3. **Position vs. Time $ x(t) $**:
- The position function is cubic:
$$
x(t) = 0.2t^3 + 4.4t + 1.4
$$
- At $ t = 1 $, $ x(1) = 6.0 \, \text{m} $ and at $ t = 2 $, $ x(2) = 11.8 \, \text{m} $.
### Graphs
While I cannot create visual graphs directly, I can describe how to sketch them:
- **Acceleration Graph**: A straight line starting from the origin (0,0) and increasing linearly to (20, 24).
- **Velocity Graph**: A parabola opening upwards, starting at (1, 5) and increasing steeply to (20, 244.4).
- **Position Graph**: A cubic curve starting at (1, 6) and increasing more steeply as $ t $ increases, reaching (2, 11.8).
### Summary of Results
- **(a)** The velocity at $ t = 20 \, \text{s} $ is $ 244.4 \, \text{m/s} $.
- **(b)** The position at $ t = 2.0 \, \text{s} $ is $ 11.8 \, \text{m} $.
- **(c)** The graphs of acceleration, velocity, and position are described above.
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