Question

At any instant t = 0 a motorbike start from rest in a given direction, a car moving with constant speed overtakes the
motorbike at that instant when it is moving with a speed 40 m/s. Motor bike accelerates uniformly till t = 18s and
then move with constant speed and overtakes the car at t = 27 s.

The maximum speed of motorbike in m/s is
A) 80
(B) 60
(C) 100
(D) 120

plz tell
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Answer 1

Answer:

60 m/sec

Explanation:

At t=0,u=0for the motorbike

And at t=18, v=18a(which will be the Vmax)

So, S1=1/2a18^2=162a..........eq (I)

And S2=(27-18)*18a=162a........eq (ii)

S1+S2= distance travelled by the car in 27 sec

S1+S2=40*27=1080.......(iii)

Putting the value of S1and S2 from eq (I) and (ii) in eq (iii)

162a+162a=1080

a =1080/324=3.33m/sec^2

Vmax=18*3.33=59.94=60m/sec.

Answer 2

Explanation:

hope it might be helpful