Question

The acceleration of a bus is given by 𝑎𝑥(𝑡) = (1.2 m/s³) t.

(a) If the bus’ velocity at time t=1.0 s is

5.0 m/s, what is its velocity at time t=2.0 s?

(b) If the bus’ position at time t=1.0 s is 6.0 m, what is

its position at time t=2.0 s?

Answer 1

(a) Acceleration is the derivative of the velocity with respect to time:

$$a=\dfrac{dv}{dt}.$$

Then, we can find the velocity of the bus by integrating $a_x(t)$ over time:

$$\displaystyle\intop_{v_0}^v dv=\displaystyle\intop_{0}^t adt,$$ $$v-v_0=1.2\displaystyle\intop_{0}^t tdt,$$ $$v=v_0+0.6t^2.$$

Let's first find $v_0$ at t=1.0 s:

$$v_0=v-0.6t^2=5\dfrac{m}{s}-0.6\cdot(1)^2=4.4\ \dfrac{m}{s}.$$

Finally, we can find velocity of the bus at t=2.0 s:

$$v(t=2\ s)=4.4\ \dfrac{m}{s}+0.6\cdot(2\ s)^2=6.8\ \dfrac{m}{s}.$$

(b) Velocity is the derivative of the position with respect to time:

$$v=\dfrac{dx}{dt}.$$

Then, we can find the position of the bus by integrating $v_x(t)$ over time:

$$\displaystyle\intop_{x_0}^x dx=\displaystyle\intop_{0}^t vdt,$$ $$x-x_0=\displaystyle\intop_{0}^t v_0dt+\displaystyle\intop_{0}^t 0.6t^2dt,$$ $$x=x_0+v_0t+0.2t^3.$$

Let's first find the inititial positon of the bus at t=1.0 s:

$$x_0=6.0\ m-4.4\ \dfrac{m}{s}\cdot1\ s-0.2\cdot(1\ s)^3=1.4\ m.$$

Finally, we can find position of the bus at t=2.0 s:

$$x(t=2\ s)=1.4\ m+4.4\ \dfrac{m}{s}\cdot2\ s+0.2\cdot(2\ s)^3=11.8\ m.$$