a car travelling at spped 54 km/he is brought to rest in 90s .find the redardation and distance travelled by the car before comming rest
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Explanation:
using 1st eqn of motion,
v= u+ at
where v is final velocity
u is initial velocity
a is acceleration and t is time.
converting initial velocity = 54 km/hr in m/sec
54*5/18= 15m/sec
final velocity is zero ( car comes to rest)
0= 15+ a*90
a=-15/90=-0.167m/s2
using 3rd eqn of motion v^2 - u^2= 2as where s is the distance
0- 15^2= 2*-0.167*s
s= 225/2*0.167=675m
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