A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s.Find acceleration and distance covered.

Given: u = 90km/h = 25m/s and v = 18km/h = 5m/s and t = 4s

Solution:

Acceleration:

we know that, $v = u + at$ or # a = $\frac{(v - u) }{ t}$

=> $\frac{(5 -25) }{ 4}$ = $-5 m/s^2$

'-' sign represents that the car is decelerating. 

Distance:

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

=>$ s = \frac{(v^2 - u^2)}{2a}$

=> $s = \frac{(5^2 - 25^2) }{ -10}$

=>$ s = 60 $metres

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Updated on: 10-Oct-2022

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