A truck is moving uniformly at a speed of 54 km/h. On seeing a ditch ahead at some distance, the driver applies breaks and decelerates the truck at the rate of 1.5 m/s² for four seconds. As a result, the truck stops just in front of the ditch.

How far away is the truck from the ditch?

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Solution

$I n i t i a l s p e e d u = 54 k m / h = 15 m / s d e c e l e r a t i o n a = - 1.5 m / s^{2} t i m e t a k e n t = 4 s d i s \tan c e t r a v e l l e d s = u t + \frac{1}{2} a t^{2} s = 15 \times 4 - 0.5 \times 1.5 \times 16 s = 48 m$

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A truck is moving uniformly at a speed of 54 km/h. On seeing a ditch ahead at some distance, the driver applies breaks and decelerates the truck at the rate of 1.5 m/s2 for four seconds. As a result, the truck stops just in front of the ditch. How far away is the truck from the ditch?

Initial speed u = 54 km/h = 15m/sdeceleration a = - 1.5 m/s2time taken t = 4 sdistance travelled s = ut + 12at2s = 15×4-0.5×1.5×16s = 48 m

A truck is moving uniformly at a speed of 54 km/h. On seeing a ditch ahead at some distance, the driver applies breaks and decelerates the truck at the rate of 1.5 m/s² for four seconds. As a result, the truck stops just in front of the ditch.

How far away is the truck from the ditch?

$I n i t i a l s p e e d u = 54 k m / h = 15 m / s d e c e l e r a t i o n a = - 1.5 m / s^{2} t i m e t a k e n t = 4 s d i s \tan c e t r a v e l l e d s = u t + \frac{1}{2} a t^{2} s = 15 \times 4 - 0.5 \times 1.5 \times 16 s = 48 m$