Question

A body moves from rest with a uniform acceleration and travels 270m in 3s. Find the velocity of the body at 10s after the start.​

Answer 1

Explanation:

u=0,s=270m,t=3s

u=0,s=270m,t=3ss=ut+ at

270=0×3+

270=0×3+

270=0×3+ ×a×3 in

a=60 m/s

Now putting the value of a in v=u+at,

Now putting the value of a in v=u+at,Velocity after t=10s is given by

Now putting the value of a in v=u+at,Velocity after t=10s is given byv=0+60×10=600 m/s

Answer 2

Answer:

HIII

Explanation:

Given :

Initial velocity (u) = 0 m/s.

Distance (s) = 270 m.

Time (t) = 3 secs.

To find :

Acceleration (a) =?

Velocity after 10 sec.

Solution :

We will use third equation of motion here;

s = ut + 1 / 2 at²

⇒ 270 = 0 * 3 + 1 / 2 * a * 3 * 3

⇒ 270 = 9 / 2 a

⇒ 9a = 540

⇒ a = 540 / 9

⇒ a = 60 m/s²

Now,

We will use first equation of motion here ;

After 10 secs.

= u + at

⇒ v = 0 + 60 * 10

⇒ v = 600

⇒ v = 600 m/s.

Hence,

The final velocity acquired by the car is 600m/s.

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