Question

an object start from rest and travels with uniform acceleration 7m in 4th second the acceleration of the object will be

Answer 1

Answer:-

Acceleration of object = 0.875 m/s²

$\rule{200}{1}$

Here,given conditions:-

• Distance (s) = 7 m
• Time (t) = 4 seconds.
• Initial velocity (u) = 0 m/s²

To find out: —

• Acceleration of object = ?

Solution:-

Here as the object starts from rest,so it's initial velocity will be 0

★ We know that,

$\star\:{\boxed{\sf\green{s = ut + \dfrac{1}{2} at^2 }}}$

Here, s = distance crossed, a = acceleration, t = time, u = initial velocity.

• Putting values:-

$\longrightarrow\sf 7 = 0(4) + \dfrac{1}{2} a(4)^2 \\\\\longrightarrow\sf 7 = \dfrac{1}{2} (a \times 16) \\\\\longrightarrow\sf 7 = \dfrac{1}{\cancel{2}}\times \cancel{16}a \\\\\longrightarrow\sf 7 = 8a \\\\\longrightarrow\sf a = 7/8 \\\\\longrightarrow\large{\green{\underline\blue{\boxed{\sf\blue{a = 0.875 \: ms^{-2} }}}}}$

Therefore,

Acceleration of object is 0.875 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.