Question

A body starts from rest with an acceleration of 2 m/s2. Calculate:
(1) velocity after 5 seconds,
(ii) the distance covered in 5 seconds,
Cho
(iii) the distance covered by the body while achieving velocity of 40 m/s.​

Answer 1

Answer:

given, a=2m/s2

u=0(as the body starts from rest)

t=5sec

so,v=u+at (here v is final velocity)

v=0+2*5

v=0+10

(1) v=10m/s

Explanation:

(2) s(distance)=ut+1/2at2

t=5sec

u=0

a=2m/s2

s=(0*5)+(1/2*2*5*5)

s=0+25

(2) s=25m

(iii)

u=0

v=40m/s

a=2m/s2

s=?

(v)2+(u)2=2as

(40)2+(0)2=2*2*s

1600+0=4*s

1600=4*s

1600/4=s

400m=s(3)

Answer 2

Explanation:

1- V-U/T = 2

V-0/5= 2

V= 2x5=10

=> V=10m/s

2-Distance- VxT=10x5= 50m

3 - Here we know that the distance covered while it has velocity 10m/s is 50m

Now new velocity will be 40 m/s

Let's take out the time it will take to achieve a velocity of 40m/s

New Initial Velocity (U)- 10m/s and V= 40m/s and acceleration (a)= 2m/s²

T= V-U/a

=>40-10/2

=>15

Therefore it will take 15 seconds to reach velocity of 40m/s

Now- Distance= Speed x Time

S= 40x15

S= 600 m

Therefore it will reach 600 m while achieving a velocity of 40m/s