Physics, asked by tanu2003, 1 year ago

a car is moving at rate of 70 kilometre per hour and applies brakes which provide a retardation of 5 metre per second square
(a)how much time does the car takes to stop
(b)how much distance does the car cover before coming to rest (c)what would be the stopping distance needed if speed of the car is doubled

Sunandit: i can answer but it will take a lot of time and efforts

Sunandit: if if they were different different questions i would definitely helped u

tanu2003: please help me

Sunandit: ok

Answers

Answered by kvnmurty

u = 70kmph= 175/9 m/s
a = -5m/s. v = 0
t = (v - u)/a = 35/9 sec
s = (v^2 - u^2)/2a
= 175*175/9*9 * 1/10 m
= 37.80 m

tanu2003: i can't understand this ans please can any other can write this question i m in 9th standard

kvnmurty: There are three formulas. Of kinematics. Know them. Also convert kmph into meters per sec.

kvnmurty: v = u + a t

kvnmurty: s = u t + 1/2 a t^2

kvnmurty: v^2 - u^2 = 2 a s

kvnmurty: Apply these formulas when there is acceleration

kvnmurty: I forgot to answer the 2nd part. If velocity u is doubled then distance s will become 2^2= 4 times.

tanu2003: this is first part

tanu2003: this ans u had answered this is "a"part

kvnmurty: I answered parts a and b above in the answer. In the comments I mentioned answer for part c.

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