Question

On turning a corner , a motorist with a speed of 90 km/h finds a child on road at 50 m ahead. He applies brakes and stops bike just in front of the child . Find retardation and time taken ?​

Answer 1

Answer:

Explanation:

initial velocity(u) = 90km/hr

=90×5/18 m/s

=25 m/s

distance(S)=50 m

final velocity(v)=0m/s(as he applies brakes)

From 3rd equation of motion,

v^2=2aS + u^2 ( where a = acceleration)

0^2=2×a×50 + 25^2

0=100a +625

-100a = 625

-a = 625/100

-a=6.25

so..retardation is 6.25m/s^2

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Answer 2

initial velocity(u) = 90km/hr

=90×5/18 m/s

=25 m/s

distance(S)=50 m

final velocity(v)=0m/s(as he applies brakes)

From 3rd equation of motion,

v^2=2aS + u^2 ( where a = acceleration)

0^2=2×a×50 + 25^2

0=100a +625

-100a = 625

-a = 625/100

-a=6.25

so..retardation is 6.25m/s^2

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