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Question: A motor car moving with a uniform speed of \(20 \mathrm {~m} / \mathrm { sec }\) comes to stop on ...

A motor car moving with a uniform speed of 20 m/sec20 \mathrm {~m} / \mathrm { sec } comes to stop on the application of brakes after travelling a distance of 10 m Its acceleration is

A

20 m/sec220 \mathrm {~m} / \mathrm { sec } ^ { 2 }

B

20m/sec2- 20 m / \mathrm { sec } ^ { 2 }

C

40m/sec2- 40 m / \sec ^ { 2 }

D

+2m/sec2+ 2 m / \sec ^ { 2 }

Answer

20m/sec2- 20 m / \mathrm { sec } ^ { 2 }

Explanation

Solution

From v2=u2+2aSv ^ { 2 } = u ^ { 2 } + 2 a S0=u2+2aS0 = u ^ { 2 } + 2 a S

a=u22S=(20)22×10=20 m/s2a = \frac { - u ^ { 2 } } { 2 S } = \frac { - ( 20 ) ^ { 2 } } { 2 \times 10 } = - 20 \mathrm {~m} / \mathrm { s } ^ { 2 }