Question

A car initially at rest starts moving with an acceleration of 4ms-2 the time in which it picks up a velocity of 72ms–1 is

Answer 1

Answer:

Hi

Intial velocity =0 m/s

Final velocity = 72m/s

Accelaration=4m/s^2

Therefore

V=u+at

72=0+(4×t)

72=4t

Therefore..... t=72÷4

T=18 secs

Explanation:

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Time\:taken=18\:sec}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about A car initially at rest starts moving with an acceleration of 4ms^2.

• We have to find Time taken to attain velocity of 72 m/s.

$\green{\underline\bold{Given :}} \\ :\implies \text{Initial\:velocity(u)=0 m/s} \\ \\ :\implies \text{Final \: velocity (v)= 72 \:m/s} \\ \\ :\implies \text{Acceleration= 4 m/ s}^{2}\\ \\ \red{\underline \bold{To \: Find : }} \\ :\implies \text{Time\:taken(t)= ?}$

• According to given question :

$\bold{Using \: first \: equation \: of \: motion : } \\ : \implies v = u + at \\ \\ : \implies 72= 0 + 4 \times t \\ \\ : \implies t \times 5 = {72} \\ \\ : \implies t= \frac{72}{4} \\ \\ \green{ : \implies {\text{t = 18\: sec}}}$