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A car starts moving with uniform acceleration \(a\) and retards uniformly at uniform deceleration \(b\).Find the average speed of the car.

Total time = \(t\)

Note by Swapnil Das
2 years, 4 months ago

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Plot a velocity time graph with two lines : one with slope \( +a \) and another with slope \( (-b) \).Then, average speed is given by the area under that graph divided by the total time interval, which can be easily obtained to be \[\frac {abt}{2 (a+b)}\]

I guess this is the simplest way.

Karthik Venkata - 2 years, 4 months ago

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Thank you very much.

Swapnil Das - 2 years, 4 months ago

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An amazing approach which you can use to solve problems of thist type and many other in kinematics is applying time reversal. For example suppose we make a video of the body which initially accelerates with acc. a and decelerates with b. Now when you will see the video backwards the body initially acc. with acc b and decelerates with a. And the time taken for both the acc and deceleration will be same as before. (If it took 1sec to acc. initially then it would take 1 sec again when decelerating in the reverse video.)

Now continuing with the problem. Suppose acc. time is t(1) and decelarating time is t (2). Then velocity attained when the particle changes its acc. is

V=a.t(1)

and from reverse video method as described earlier it is the same v in opposite direction. V=b.t(2) So we get

at(1)=bt(2) and t(1)+t(2)= T

\[t_{1}=\frac {b}{a}t_{2}\] \[(1+ b/a)t_{2}=T\] Now we can determine t1 and t2 very easily. Average velocity is total distance / total time. Determine distance by using second equation of motion for the two time intervals t1 and t2 separately. The answer should come out to be \[\frac {abT}{2 (a+b)}\]

The approach is more useful in many other questions.

Satvik Choudhary - 2 years, 4 months ago

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