Question

An object starts from rest and travels 7m with uniform acceleration in 4th second.. What will be the acceleration of the object?

Answer 1

Given that, the object starts from rest means its initial velocity (u) from where it starts is 0 m/s. And it covers a distance (s) of 7 m.

Also, time (t) is 4 seconds.

We have to find the acceleration of the object.

Using Newton's Second equation of Motion

s= ut + 1/2 at²

Substitute the known values

→ 7 = 0(4) + 1/2 × a × (4)²

→ 7 = 0 + 1/2 × 16a

→ 7 = 1/2 × 2(8a)

→ 7 = 8a

→ 7/8 = a

→ 0.87 = a

Therefore, the acceleration of the object is 0.87 m/s²

Answer 2

☯ GiveN :

Initial velocity of object (u) = 0m/s (∵ It starts from rest).

Distance covered (s) = 7 cm

Time taken (t) = 4 seconds.

$\rule{200}{1}$

☯ To FinD :

We have ro find the acceleration done by the object.

$\rule{200}{1}$

☯ SolutioN :

We know that,

$\Large{\implies{\boxed{\boxed{\sf{s = ut + \frac{1}{2} \: at^2}}}}}$

★ Putting Values ★

$\sf{\dashrightarrow 7 = 0(4) + \frac{1}{2} \: a \times (4)^2} \\ \\ \sf{\dashrightarrow 7 = 0 + \frac{1}{\cancel{2}} \times \cancel{16}} \\ \\ \sf{\dashrightarrow 7 = 8a} \\ \\ \sf{\dashrightarrow a = \frac{7}{8}} \\ \\ \sf{\dashrightarrow a = 0.875} \\ \\ \Large{\implies{\boxed{\boxed{\sf{a = 0.875 \: ms^{-2}}}}}}$

$\therefore$ Acceleration done by object is 0.875 m/s².