The speedometer readings of a car are shown below. Find the acceleration of the car and its displacement.

Time	Speedometer
9:25 am	36 km/h
9:45 am	72 km/h

Given:

Time, $t$ = 20 min = 20 $\times$ 60 seconds = 1200 sec  $[\because the\ difference\ between\ the\ time\ 9:45\ and\ 9:25\ is\ 20]$

Initial velocity, $u$ = 36 km/h = $36\times {\frac {5}{18}}$ = 10 m/s  [changed km/h into m/s by multiplying $\frac {5}{18}$]

Final velocity, $v$ = 72 km/h = $72\times {\frac {5}{18}}$ = 20 m/s    [changed km/h into m/s by multiplying $\frac {5}{18}$]

To find: Acceleration, $a$ of the car and its displacement $s$. 

Solution:

We know that-

$a=\frac {v-u}{t}$

Subtituting the given values in the formula we get-

$a=\frac {20-10}{1200}$

$a=\frac {10}{1200}$

$a=\frac {1}{120}$

$a=0.008m/s^2$

Thus, the acceleration $a$ of the car will be 0.008 $m/s^2$

Now,

To find the displacement of the car, we shall use the 2rd equation of motion.

From Second equation of motion we have:

$v^2-u^2=2as$

$(20)^2-(10)^2=2\times({\frac {1}{120}})s$

$400-100=(\frac {1}{60})s$

$300=\frac {1}{60}s$

$s=300\times {60}$

$s=300\times {60}$

$s=18000m\ or\ 18km$

Thus, the magnitude of the displacement or distance will be 1800m or 18km.