Question

(11) A body starting from rest has a uniform acceleration of 20 cm s-2.
What is its velocity after moving 40 cm ? How long does it take to travel this
distance ?​

Answer 1

u = 0

v= ?

a = 20 cm s^-2

s = 40 cm

v^2 = u^2 + 2as

v^2 = 0 +2×20×40

v^2 = 40×40

v = 40 m s^-1

v = u+at

40 = 0 + 20 t

40 = 20 t

t = 2 s

Answer 2

Here,

the initial velocity, u=0 cm/s

the distance, s= 40 cm

the acceleration, a = 20 cm s‐² or 20 cm/s²

the velocity , v = ?

the time taken , t=?

now, s=ut+½at² from this equation we get,

or, 40cm =0×t+½×(20 cm/s²) × t²

or, 40cm =10 cm/s² × t²

or, 40cm/10 cm/s² = t²

or, t² =40cm×s²/10cm

or, t² = 4 × s²

or, t = 2×2×s×s since this equation by square root over.

or, t = 2a

therefore time taken,t = 2s and

the velocity, v = distance/ time or (s/t)

v= 40 cm / 2s

v = 20 cm/s

therefore the velocity is 20 cm/s.