Question

The motion of cars 'A' and 'B' is described by the figure shown below: car 'A' moves with a constant velocity V_A = 9.5 m/s and car 'B' speeds up from rest with a constant acceleration rate so it reaches the speed of car 'A' at t = 5 seconds.

The acceleration of car 'B', a_B =  

If initially (at t = 0 ) the cars were at the same position, at what distance relative to the initial point will the cars meet again?

The distance where the cars meet, D =  

When will the cars meet?

The time of the meeting, t =

At the time of meeting, what will be the speed of car 'B'?

The speed of car 'B', V_B =  

Answer 1

Velocity of car A is V_A = 9.5 m/s

Since Car B starting from rest with uniform accelaration a_B reach velocity of 9.5 m/s at t = 5 s

So 9.5 = 0 + a_B (5)

or, a_B = 1.9 m/ s²

^{Let the cars meet again after time t s by travellin a ditance of D m, then}

^{For car A , D = 9.5t ----------(i )}

For car B, we have D = 0.5 (1.9 ) t² ------------ ( ii )

From both the relation we can say

0.5 (1.9 ) t² = 9.5 t

or, 0.95 t ( t - 10 ) = 0

or, t -10 = 0 ( since t = 0 represent initial condition )

or, t = 10 s  

# So D = 9.5 ( 10 ) m

= 95 m

# The time of meeting t = 10 s

# Speed of car B at the ime of meeting V_B = 0 + a_B t

= 1.9 ( 10 ) m/s

= 19 m/s