An object travels north with a velocity of 10 ms- 1 and then speeds up to a velocity of 25 ms-1 in 5 s. The acceleration of the object in these 5 s is

Question

An object travels north with a velocity of 10 ms- 1 and then speeds up to a velocity of 25 ms-1 in 5 s. The acceleration of the object in these 5 s is (A) 12 m s -2 in north direction (B) 3 m s -2 in north direction (C) 15 m s -2 in north direction (D) 3 m s -2 in north direction

Tardigrade Admin · Accepted Answer

Acceleration a =(Δ t/Δ t)=(v2 -v1/Δ t) Given v, = 10 ms-1 ,v2 = 25 ms-1, Δ t = 5s hence a=(25-10/5) 3ms-2in north direction.

Q. An object travels north with a velocity of $10 m s^{- 1}$ and then speeds up to a velocity of $25 m s^{- 1}$ in 5 s. The acceleration of the object in these 5 s is

$12 m$ $s^{- 2}$ in north direction

$3 m$ $s^{- 2}$ in north direction

$15 m$ $s^{- 2}$ in north direction

$3 m$ $s^{- 2}$ in north direction

Acceleration a = $\frac{Δ t}{Δ t} = \frac{v _{2} - v _{1}}{Δ t}$
Given v, = $10 m s^{- 1}, v_{2} = 25 m s^{- 1}, Δ t = 5 s$
hence $a = \frac{25 - 10}{5} 3 m s^{- 2}$ in north direction.

Q. An object travels north with a velocity of 10ms−1 and then speeds up to a velocity of 25ms−1 in 5 s. The acceleration of the object in these 5 s is

12m s−2 in north direction

3m s−2 in north direction

15m s−2 in north direction

3m s−2 in north direction

Acceleration a =ΔtΔt​=Δtv2​−v1​​ Given v, = 10ms−1,v2​=25ms−1,Δt=5s hence a=525−10​3ms−2in north direction.

Q. An object travels north with a velocity of $10 m s^{- 1}$ and then speeds up to a velocity of $25 m s^{- 1}$ in 5 s. The acceleration of the object in these 5 s is

$12 m$ $s^{- 2}$ in north direction

$3 m$ $s^{- 2}$ in north direction

$15 m$ $s^{- 2}$ in north direction

$3 m$ $s^{- 2}$ in north direction

Acceleration a = $\frac{Δ t}{Δ t} = \frac{v _{2} - v _{1}}{Δ t}$
Given v, = $10 m s^{- 1}, v_{2} = 25 m s^{- 1}, Δ t = 5 s$
hence $a = \frac{25 - 10}{5} 3 m s^{- 2}$ in north direction.