Question

A car initially at rest start moving with acceleration 0.5 ms^-2 covers a distance of 25m. calculate the time required to cover this distance and the final velocity of the car.​

Answer 1

Answer:

S=ut+1/2at^2

where s=distance travelled

u=initial velocity

a=acceleration

t=time

ATQ, u=0m/s(as the body starts from rest, thus it's obvious that the initial velocity will be 0),t=20s, a=5m/s^2, s=?

Now put the value in the formula:

s=ut+1/2at^2

s=0(20s)+ 1/2(5m/s^2)(20s)^2

s=0+1/2(5m/s^2)(400s^2)

s=1/2(2000m)

s=1000m

Thus, distance travelled is 1000m

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Answer 2

Answer:

initial velocity(u)=0

acceleration=0.5m/sec^2

distance(s)=25m

t=?

final velocity(v)=?

v^2=u^2+2as

(0)^2=u^2+2(0.5)(25)

(0)^2=u^2+25

-25=u^2

u=+-5m/sec

v=u+at

0=5+0.5*t

-5=0.5t

-5*10/5=t

t=10sec

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