A driver, driving a truck of mass 10 t at a velocity of ${72 k m h r}^{- 1}$ , sees an obstruction on the road at a distance of 145 m. If the personal reaction of driver is 0.8 s and retarding force of brakes 16000N, find whether accident takes place or not ?

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Solution

We have,

$m a s s = 10 t = 1000 k g$

$I n i t i a l d i s t a n c e = 145 m$

$V e l o c i t y = 72 k m / h r = 72 \times \frac{5}{18} = 20 m / s$

$R e t a r d a t i o n f o r c e = 1600 N$

So, $a = \frac{16000}{10000} = 1.6 m / s^{2}$

Now,

Initial distance is 145m, After 0.8 s the minimum distance is:

$145 = 0.8 \times 20$

$= 145 - 16 = 129 m$

Again, We know that,

Let D is the minimum gap left

$v^{2} = U^{2} + 2 a s$

$D - 400 = 1.65 \times 2$

$s = \frac{400}{3.2} = 125 m$

So,

$129 - 125 = 4 m$

Thus, The minimum gap is $4 m$

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A driver, driving a truck of mass 10 t at a velocity of 72kmhr−1, sees an obstruction on the road at a distance of 145 m. If the personal reaction of driver is 0.8 s and retarding force of brakes 16000N, find whether accident takes place or not ?

A driver, driving a truck of mass 10 t at a velocity of ${72 k m h r}^{- 1}$ , sees an obstruction on the road at a distance of 145 m. If the personal reaction of driver is 0.8 s and retarding force of brakes 16000N, find whether accident takes place or not ?