A tractor of 3,000 kg, moving at a constant speed of 20 m/s stops after covering some distance. The force applied by the brakes is 24,000 N. Compute the time taken by the truck to stop after the brake is applied.

-2.5 s

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6 s

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5 s

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2.5 s

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Solution

The correct option is D 2.5 s
Given, m = 3,000 kg, F = 24,000 N, u = 20 m/s, v = 0 m/s (stops)
a = F/ m = - (24,000)/3,000 = -8 $m / s^{- 2}$ [Negative sign because of retardation, truck stops, velocity decreases]
Newton’s 1st law of motion v = u + at
0 = 20 + (-8 $m / s^{- 2}$ ) t
t = 2.5 s

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A tractor of 3,000 kg, moving at a constant speed of 20 m/s stops after covering some distance. The force applied by the brakes is 24,000 N. Compute the time taken by the truck to stop after the brake is applied.​

The correct option is D 2.5 sGiven, m = 3,000 kg, F = 24,000 N, u = 20 m/s, v = 0 m/s (stops) ​ a = F/ m = - (24,000)/3,000 = -8 m/s−2 [Negative sign because of retardation, truck stops, velocity decreases]​ Newton’s 1st law of motion v = u + at​ 0 = 20 + (-8 m/s−2) t​ t = 2.5 s ​

A tractor of 3,000 kg, moving at a constant speed of 20 m/s stops after covering some distance. The force applied by the brakes is 24,000 N. Compute the time taken by the truck to stop after the brake is applied.

The correct option is D 2.5 s
Given, m = 3,000 kg, F = 24,000 N, u = 20 m/s, v = 0 m/s (stops)
a = F/ m = - (24,000)/3,000 = -8 $m / s^{- 2}$ [Negative sign because of retardation, truck stops, velocity decreases]
Newton’s 1st law of motion v = u + at
0 = 20 + (-8 $m / s^{- 2}$ ) t
t = 2.5 s