A car moving in a straight line starts at$x = \mathbf{0}$at$\mathbf{t} = \mathbf{0}$. It passes the point$\mathbf{x} = \mathbf{25}\mathbf{.}\mathbf{0} \mathbf{m}$with a speed of 11.0 m/s at$\mathbf{t} = \mathbf{3}\mathbf{.}\mathbf{0} \mathbf{s}$. It passes the point$\mathbf{x} = \mathbf{385} \mathbf{m}$with a speed of 45.0 m/s at$\mathbf{t} = \mathbf{20}\mathbf{.}\mathbf{0} \mathbf{s}$. Find (a) the average velocity and (b) the average acceleration, between$\mathbf{t} = \mathbf{3}\mathbf{.}\mathbf{0} \mathbf{s}$and$\mathbf{t} = \mathbf{20}\mathbf{.}\mathbf{0} \mathbf{s}$.

The average velocity of an object is known as the average distance traveled by a particular object in a particular amount of time.

Given data.

At $t_1 = 0$, the car is at $x_1 = 0$.

At $t_2 = 3.0 \mathrm{s}$, the car is at $x_2 = 25.0 \mathrm{m}$and is moving with a speed of $v_2 = 11.0 \frac{\mathrm{m}}{\mathrm{s}}$.

At $t_3 = 20.0 \mathrm{s}$, the car is at $x_3 = 385 \mathrm{m}$and is moving with a speed of $v_3 = 45.0 \frac{\mathrm{m}}{\mathrm{s}}$.

Rule.

You know that

$\begin{array}{c}\mathrm{Average}\mathrm{velocity} = \frac{\mathrm{Total}\mathrm{displacement}}{\mathrm{Total}\mathrm{time}}\\ = \frac{\mathrm{Finalposition}- \mathrm{Initialposition}}{\mathrm{Total}\mathrm{time}}\\ \mathrm{Average}\mathrm{accelaration} = \frac{\mathrm{Change}\mathrm{in}\mathrm{velocity}}{\mathrm{Total}\mathrm{time}}\\ = \frac{\mathrm{Final}\mathrm{velocity} - \mathrm{Initial}\mathrm{velocity}}{\mathrm{Total}\mathrm{time}}\end{array}$

(a) Now, the average velocity between $t_2 = 3.0 \mathrm{s}$and $t_3 = 20.0 \mathrm{s}$is

$\begin{array}{c}v = \frac{x_3- x_2}{t_3- t_2}\\ = \frac{385 \mathrm{m}- 25.0 \mathrm{m}}{20.0 \mathrm{s}- 3.0 \mathrm{s}}\\ = 21.2 \frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Therefore, the average velocity is $21.2 \frac{\mathrm{m}}{\mathrm{s}}$between $t_2 = 3.0 \mathrm{s}$and $t_3 = 20.0 \mathrm{s}$.

(b) Now, the average acceleration between $t_2 = 3.0 \mathrm{s}$and $t_3 = 20.0 \mathrm{s}$is

$\begin{array}{c}a = \frac{v_3- v_2}{t_3-t_2}\\ = \frac{45.0 \mathrm{m}- 11.0 \mathrm{m}}{20.0 \mathrm{s}- 3.0 \mathrm{s}}\\ = 2.00 \frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Therefore, the average acceleration is $2.00 \frac{\mathrm{m}}{{\mathrm{s}}^2}$between $t_2 = 3.0 \mathrm{s}$and $t_3 = 20.0 \mathrm{s}$.