Question

on turning a corner a motorist with a speed of 90 km per hour find a child on the road 50meter ahead . he applies brakes and stop the car just in front of children calculate the retardation​

Answer 1

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

Answer 2

Answer:

since u = 90 km per hour = 25 m per second

therefore, v²=u²+2as

0² = 25² + 2 * a * 50

0 = 625 + 100a

-625 = 100a

a = -625/100

a = - 6.25 m per s²

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