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
if if they were different different questions i would definitely helped u
please help me
ok
Answers
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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
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
i can't understand this ans please can any other can write this question i m in 9th standard
There are three formulas. Of kinematics. Know them. Also convert kmph into meters per sec.
v = u + a t
s = u t + 1/2 a t^2
v^2 - u^2 = 2 a s
Apply these formulas when there is acceleration
I forgot to answer the 2nd part. If velocity u is doubled then distance s will become 2^2= 4 times.
this is first part
this ans u had answered this is "a"part
I answered parts a and b above in the answer. In the comments I mentioned answer for part c.
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