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

Total time = \(t\)

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

Total time = \(t\)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestPlot 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 \[\frac {abt}{2 (a+b)}\]

– Karthik Venkata · 1 year, 11 months agoLog in to reply

– Swapnil Das · 1 year, 11 months ago

Thank you very much.Log in to reply

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 · 1 year, 11 months ago

Log in to reply