Question

A body starts from rest and travels with a unifom acceleration a to make a displacement is 6m/s then its unifom acceleration is​

Answer 1

Correct Question !!

A body starts from rest and travels with a unifom acceleration a to make a displacement of 6m and its final velocity is 6m/s then its unifom acceleration is ?

Given:-

• Initial velocity ,u = 0m/s
• Final velocity ,v = 6m/s
• Displacement,s = 6m

To Find:-

• Acceleration ,a

Solution:-

⠀⠀⠀⠀⠀⠀⠀⠀According to the Question

It is given that the body starts from rest & cover a displacement of 6 m with velocity 6m/s. We calculate the acceleration of the body . Using 3rd equation of motion

• v² = u² + 2as

where,

• v denote final velocity
• u denote initial velocity
• a denote acceleration
• s denote displacement

Substitute the value we get

$:\implies$ 6² = 0² + 2×a × 6

$:\implies$ 36 = 0 + 12a

$:\implies$ 36 = 12a

$:\implies$ 12a = 36

$:\implies$ a = 36/12

$:\implies$ a = 3m/s²

• Hence, the uniform acceleration of the body is 3m/s².

Answer 2

Appropriate Question:-

A body starts from rest and travels with a unifom acceleration a to make a displacement is 6m with velocity of 6m/s. Then its unifom acceleration??

Given:-

• A body starts from rest and travels with a unifom acceleration a to make a displacement is 6m/s.

To Find:-

• It's unifom acceleration

Equation used:-

• ${\boxed{\bf{v^2=2as+u^2}}}$

Here,

• v = Final Velocity

• u = Initial Velocity

• a = Acceleration

• S = Displacement

Solution:-

Using, v² = 2as + u²

Here,

• v = 6m/s

• S = 6m

• u = 0m/s

Putting Values,

$\sf \implies\:6^2=2\times a \times 6+0^2$

$\sf \implies\:12a=36$

$\sf \implies\:a=\dfrac{36}{12}$

$\sf \implies\:a=\dfrac{9}{3}$

$\bf \implies\:a=3ms^{-2}$

Hence, It's Unifom Acceleration is 3m/s².

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More Equations of Motion:-

• $\bf v=u+at$

• $\bf S=ut+\dfrac{1}{2}at^2$

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