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initially car a is 10.5m ahead of b.both start moving at time t=0 in the same direction along a straight line.the velocity time graph of two cars is shown in figure.the time the car b will catch car a,will be

initially car a is 10.5m ahead of b.both start moving at time t=0 in the same direction along a straight line.the velocity time graph of two cars is shown in figure.the time the car b will catch car a,will be

Grade:12th pass

2 Answers

Blessy
11 Points
6 years ago
Let car B will catch car A after travelling a distance of 10.5 + x.time taken by both the car to meet eachother will be same.Therefore the eqation will bevt+10.5=1/2at^2where v=10m/s and a=1m/s^2
Ayush
15 Points
4 years ago
We know car A is 10.5m ahead of car B. This is implies that for this case:
               S=ut+\tiny \frac{1}{2}at2+10.5.
                  =10t+10.5
And for 2nd case:
               S=ut+\tiny \frac{1}{2}at2
                  =\tiny \frac{1}{2}t2    [Since, slope is 1]
Now ,equating both the cases we would get:
t=21 sec.

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