FrictionHard
Question
A car starting from rest accelerates at the rate f through a distance s, then continues at constant speed for time t and then decelerates at the rate
to come to rest. If the total distance travelled is 15 s, then :-
Options
A.s = ft
B.s =
ft2
C.s =
ft2
D.None of these
Solution

Let the car takes time t1 to reach from O to A a distance s with acceleration f and t2 to come to rest from B to C with deceleration
velocity at A = velocity at B
f t1 =
or t2 = 2t1
Now, s1 = 0 +
s2 = (f t1)t (∵ v = 0 + f t1)
s3 = (f t1) t2 -
= f t1 . 2t1 -
= 2f t12 - f t12 = f t12
But, s1 + s2 + s3 = 15 s
∴
ft1t = 12 s = 12
or t1 =
∴ s =
Thus, none of the option is correct.
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