Breaks applied to a car produce uniform retardation of 0.9m/ssquare if the car was travelling with the velocity 27m/s what distance will it cover before coming to rest?
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Answered by
94
When the brakes are applied to a car produced a uniform retardation of 0.9 m/s2 if the car was travelling with velocity of 27 m/s:
v = 0 m/s
u = 27 m/s
a = -0.9 m/s2
To find out the distance travelled by the car, S before coming to rest we will use Newton's third equation of motion
v2 - u2 = 2aS, so 0^2-27^2=2*-0.9*S, so, 0-729=-1.8*S,or ,-729/-1.8=S,or, S=405 km.
v = 0 m/s
u = 27 m/s
a = -0.9 m/s2
To find out the distance travelled by the car, S before coming to rest we will use Newton's third equation of motion
v2 - u2 = 2aS, so 0^2-27^2=2*-0.9*S, so, 0-729=-1.8*S,or ,-729/-1.8=S,or, S=405 km.
Answered by
136
Given : a = -0.9m/s² (because its retardation)
v = 0 (Because in the final, the car stops)
u = 27m/s (its already given)
To find : Distance 'd'
Solution : The formula we will be using is -
V²=U²+2as
So,
(0)²= (27)²+2(-0.9)(s)
0 = 729 +( -1.8)(s)
=> 1.8s = 729
=> s = 729/1.8
So, the answer is s = 405 metres!!
v = 0 (Because in the final, the car stops)
u = 27m/s (its already given)
To find : Distance 'd'
Solution : The formula we will be using is -
V²=U²+2as
So,
(0)²= (27)²+2(-0.9)(s)
0 = 729 +( -1.8)(s)
=> 1.8s = 729
=> s = 729/1.8
So, the answer is s = 405 metres!!
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