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	54 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, $v$ = 54 km/h = $54\times {\frac {5}{18}}$ = 15 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-

$a=\frac {20-15}{1200}$

$a=\frac {5}{1200}$

$a=\frac {1}{240}$

$a=0.004m/s^2$

Thus, the acceleration $a$ of the car will be 0.004 $m/s^2$

Now,

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

From Second equation of motion we have:

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

$(20)^2-(15)^2=2\times({\frac {1}{240}})s$

$400-225=(\frac {1}{120})s$

$175=\frac {1}{120}s$

$s=175\times {120}$

$s=300\times {60}$

$s=21000m\ or\ 21km$

Thus, the magnitude of the displacement or distance will be 21000m or 21km.

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Updated on: 10-Oct-2022

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