Question

A cheetah can accelerate from rest at a rate of 4ms^-2 ,(A) What will be the velocity attained by it in 10 seconds ,(B) How far will it travel in this duration .​

Answer 1

Answer:

(A)

u (Initial velocity )=0m/s  

v (Final velocity =?  

a (Acceleration)= 4m/s^2  

t (Time )= 10sec

Using first equation of motion

v = u + at

v = 0 +(4 X 10)

v= 0+40

v = 40m/s

(B)

u (Initial velocity ) = 0m.s

a (Acceleration) = 4m/s^2

t (Time) = 10 secs

s (Distance) = ?

Using second equation of motion

s = ut + (1/2) at²

s= 0 X 10 + (1/2 X 4 X 10 X10)

s= 0 + 1/2 X  400

s= 0+ 200

s= 200 m

Therefore Ans a) VELOCITY ATTAINED IN 10 SECS = 40M/S

                 Ans b) DISTANCE TRAVELLED = 200 m