A car is moving with a uniform velocity of 20 meter per second. It is brought to rest in 5 s by the application of brakes. What is the distance travelled by the car during this time interval ?

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Solution

Dear Student
$u = 20 m / s v = 0 m / s t = 5 s v = u + a t a = \frac{v - u}{t} = \frac{0 - 20}{5} = - 4 m / s^{2} N e g a t i v e s i g n s h o w s t h a t s p e e d i s d e c r e a \sin g t h a t r e t a r d a t i o n s = u t + \frac{1}{2} a t^{2} s = 20 \times 5 + \frac{1}{2} \times (- 4) \times 5 \times 5 s = 100 - 50 = 50 m$

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A car is moving with a uniform velocity of 20 meter per second. It is brought to rest in 5 s by the application of brakes. What is the distance travelled by the car during this time interval ?

Dear Student u = 20 m/sv =0 m/st = 5sv = u +ata =v-ut =0-205 =- 4 m/s2Negative sign shows that speed is decreasing that retardations= ut +12at2s =20×5 +12×-4×5×5s =100-50 = 50 m Regards