SOLUTION: 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 b

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.
RELATED QUESTIONS
A cat walks in a straight line, which we shall call the x-axis, with the positive... (answered by textot)

The acceleration of a point is given by a=4−t^2 m/s^2. Write an equation for the... (answered by Fombitz)

A particle travelling in a straight line passes a fixed point O with a velocity of 1... (answered by mananth)

Please can you answer this question A bus travels uniformly from 54 km/h to 72 km/h in (answered by Alan3354)

If distance �s� is a function of time �t� given by s = t3 � 6t2 � 15t + 12 then the... (answered by amarjeeth123)

The formula that describes an object’s motion is given by S=ut+12at2 S = u t + 1 (answered by Boreal,ikleyn)

A particles travel in a straight line so that,t seconds after leaving a fixed point,its... (answered by MowMow,ikleyn)

The velocity of an object as a function of time is V(x)= 8t-5. Calculate the acceleration (answered by ikleyn)

A particle moves in a straight line so that its distance, s m, from a fixed point A on... (answered by Alan3354)