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 rateto 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 = t2
or    t2 = 2t1
Now, s1 = 0 + ft12 = ft12 = s   (given)
s2 = (f t1)t   (∵ v = 0 + f t1)
s3 = (f  t1) t2 -
= f t1 . 2t1 - (2t1)2
= 2f t12 - f t12 = f t12
But,  s1 + s2 + s3 = 15 s
∴   f t12 + f t1t + f t12 = 15s
f t12 + f t1t = 15s
ft1t = 12 s = 12f t12
or    t1 =
∴   s =
Thus, none of the option is correct.

Create a free account to view solution

View Solution Free
Topic: Friction·Practice all Friction questions

More Friction Questions