# A motor car moving at a speed of 72km/h can not come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s.

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A motor car moving at a speed of 72km/h can not come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a higway the car is behind the truck both moving at 72km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is 0.5s.

(Comment : This is to illustrate why vehicles carry the message on the rear side. “Keep safe Distance”)

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Car behind the truck

Regardation of truck = 20/5 = 4ms-2

Let the truck be at a distance x from the car when breaks are applied

Distance of truck from A at t > 0.5 s is          x + 20t - 2t2

Distance of car from A is            10 + 20(t - 0.5) -(10/3)(t - 0.5)2

If the two meet Second method: This method does not require the use of calculus. If the car is behind the truck

Vcar = 20 – (20/3)(t – 0.5) for t > 0.5 s as car declerate only after 0.5 s.

Vtruck = 20 – 4t

Find t from equating the two or from velocity vs time graph. This yields t = 5/4 s.

In this time truck would travel truck,

Struck= 20(5/4) – (1/2)(4)(5/4)2 = 21.875m

and car would travel, Scar = 20(0.5) + 20(5/4 – 0.5) – (1/2)(20/3) x (5/4 - 0.5)2 = 23.125m

Thus Scar – Struck = 1.25m.

If the car maintains this distance initially, its speed after 1.25s will he always less than that of truck and hence collision never occurs.

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