Question

A car starting from the rest under constant acceleration attains a spped of 144 km/hr after 20 sec. The distance travelled by car is

Answer 1

Initial velocity= 0

Final velocity = 144 km/hr = 144×1000/60×60= 40 m/s (converting in the si unit)

t=20 sec

So the acceleration= final velocity-i initial velocity/ time =v-u/t=40–0/20=2m/s^2

We are going to use 2nd equation of motion , which is

s=ut+1/2at^2

Substituting the values

We get

s=0×20 + 1/2×2×20^2

s=20^2

Therefore

s = 400m ( this is the distance it will travel )

Answer 2

$\small{\underline{\tt{\green{Given-}}}}$

Initial speed u = 0

final speed v = 144 km/hr  = 144 × (5/18) = 40 m/s

Find-

distance covered in 20 s

$\huge{\underline{\tt{\blue{Solution-}}}}$

Acceleration" v = u + a t "  i.e. a = v/t = 40/20 = 2 m/s2

distance covered in 20 s :- S = u t + (1/2) a t2 = (1/2)×2×20×20 = 400 m