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  • A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3 s while for a truck this time interval is 5 s. On a highway the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency.

A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3 s while for a truck this time interval is 5 s. On a highway the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at an emergency. At what distance the car should be from the truck so that it does not bump onto the truck. Human response time is 0.5 s.

Answers (1)

Given: (for truck)

U=72\; km/hr=2\times\frac{5}{18}m/s

     =20\; m/s

V = 0, t = 5s

a =?

We know that.

V= u + at

i.e., 0 = 20 + a (5)

thus, a = \frac{-20}{5} = \frac{-4m}{s}

Given : (for car)

U = 20 \; m/s

V =0, t = 3s

a = a_{c}

Again, v = u + at

0 = 20 + a{_{c}}^{3}

a_c = \frac{-20}{3} \; m/s^2

A human takes at least 0.5 seconds to respond; thus, the time taken by the car driver to respond is (t-0.5) sec …. (car takes t time to stop)

Vc = u + a{_{c}}{t}

0 = 20-\frac{20}{3}.(t - 0.5) \; \; \; \; \; \; .......... (i)

There is no responding time for the truck driver so he applies breaks with passing signal to car's back side; hence,

V = u + at

0 = 20 - 4t \; \; \; \; \; \; .......... (ii)

From (i) & (ii),

20 - 4t = 20 - \frac{20}{3} ( t - 0.5)

-4t = - \frac{-20}{3} ( t - 0.5)

\\12t = 20t - 10\\-20 + 12t = -10\\

Thus, -8t = -10

Thus, t=\frac{10}{8}

              =1.25\; sec

Now the distance covered by the car & the truck in \frac{5}{4} sec will be,

S = 20 \left ( \frac{5}{4} \right ) + 0.5(-4) \left ( \frac{5}{4} \right )\left ( \frac{5}{4} \right )

……. ( because \; s = ut + \frac{1}{2} at^{2})

    = 25 - 3.125

     =21.875 m

In the first 0.5 seconds, the car moves at a uniform speed, but after responding, the brakes are applied for 0.5 seconds, and the retarding motion of the car starts.

S_{c} = (20 \times 0.5) \times 20 (1.25 - 0.5) + \frac{1}{2} \left ( \frac{-20}{3} \right ) (1.25 - 0.5)^{2}

       = 25 - 1.875

       =21.875 m

Thus, s_{c} - s = 23.125 - 21.875

                       = 1.25 m

Therefore, the car must be 1.25 metres behind the truck to avoid bumping into it.

Posted by

infoexpert22

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