Question

A car accelerates uniformly from 18 km/h to 36 km/h in 5 seconds. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.

Answer 1

Step 1:Given data:

Initial velocity, u = 18 km/h.

final velocity, v = 36 km/h.

time of travel, t = 5 sec.

Step 2:Finding the acceleration covered by the car:

Let, the acceleration of the car is $am/s^2.$

Now,

$$\begin{gathered} initialvelocity,u=18k/h=\frac{18\times 1000}{3600}=5m/s. \\ finalvelocity,v=36km/h=\frac{36\times 1000}{3600}=10m/s. \end{gathered}$$

We know that,

From the first equation of motion,

$v=u+at$

Where u and v are the initial and final velocities respectively, a is a constant acceleration.

$$\begin{gathered} ora=\frac{v-u}{t}=\frac{10-5}{5}=1m/s^2. \\ ora=1m/s^2. \end{gathered}$$

Therefore, the acceleration of the car is $1m/s^2.$

Step 3:Finding the distance covered by the car:

Let, the distance traveled by car is ‘S’ m.

We know that,

From the second equation of motion,

$s=ut+\frac{1}{2}a{t}^2.$

Where, u = initial velocity, t = time, s is the distance traveled, and a = acceleration.

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2=\left(5\times 5\right)+\frac{1}{2}\times 1\times \left(5^2\right) \\ ors=25+\frac{25}{2}=37.5m. \\ ors=37.5m. \end{gathered}$$

Therefore, the distance traveled by car is 37.5 m.