The speed-times relation of a car whose weight is $1500 k g$ as shown in the given graph. How much breaking force has been applied at the end of $7$ sec, to stop the car in $2$ sec?

2000 N

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1200 N

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4800 N

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8400 N

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9000 N

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Solution

The correct option is E $9000 N$
Speed - time graph is given,
Slope of the speed-time graph gives an acceleration of the moving object.

In between 7th and 9th second,
Slope of line $= \frac{0 - 12}{9 - 7} = - 6 m / s^{2}$
Slope of line is negative which show deacceleration of object. Hence When driver will apply break, car will deaccelerate with $- 6 m / s^{2}$ .
Breaking force, $F = m a = 1500 \times 6 = 9000 N$

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The speed-times relation of a car whose weight is 1500kg as shown in the given graph. How much breaking force has been applied at the end of 7sec, to stop the car in 2 sec?

The speed-times relation of a car whose weight is $1500 k g$ as shown in the given graph. How much breaking force has been applied at the end of $7$ sec, to stop the car in $2$ sec?