Question

a body starting from rest moving with a constant acceleration covers 15m in 3sec. find the distance covered by it in 12sec​

Answer 1

Given:-

• Initial velocity ,u = 0m/s
• Distance covered ,s = 15m
• Time taken ,t = 3 s

To Find:-

• Distance covered by it in 12 second.

Solution:-

It is given that the body starts from rest and moving with a constant acceleration . We have to calculate the distance covered by it in 12 second . Firstly we calculate the acceleration of the body . Using 2nd equation of motion

• s = ut + 1/2at²

where,

• s denote distance covered
• u denote initial velocity
• t denote time taken
• a denote acceleration

Substitute the value we get

$:\implies$ 15 = 0×3 + 1/2×a × 3²

$:\implies$ 15 = 0 + 1/2 × a × 9

$:\implies$ 15×2 = 9 × a

$:\implies$ 30 = 9× a

$:\implies$ a = 30/9

$:\implies$ a = 10/3 m/s² = 3.33 m/s²

Now, calculating the distance covered by body in 12 second . Again using 2nd equation of motion

• s = ut + 1/2 at²

substitute the value we get

$:\implies$ s = 0 × 12 + 1/2 × 10/3 × 12²

$:\implies$ s = 0 + 5/3 × 144

$:\implies$ s = 5 × 48

$:\implies$ s = 240 m

• Hence the distance covered by the body in 12 second is 240 metres.

Answer 2

Given:-

• A body starting from rest moving with a constant acceleration covers 15 m in 3 sec.

To Find:-

• Distance covered by it in 12 sec.

Law used:-

• ${\boxed{\bf{Second\:Law\:of\: Motion: S=ut+\dfrac{1}{2}at^2}}}$

Here,

• S = Distance

• u = Initial Velocity

• a = Acceleration

• t = Time Taken

Solution:-

Using the Second Law of Motion,

$\bf \implies\:S=ut+\dfrac{1}{2}at^2$

Here,

• u = 0

• S = 15m

• t = 3 sec

Putting the values,

$\sf \implies\:15=0\times3+\dfrac{1}{2}\times a \times 3^2$

$\sf \implies\:15=\dfrac{1}{2}\times a \times 9$

$\sf \implies\:9a=15\times2$

$\sf \implies\:a=\dfrac{30}{9}$

${\boxed{\bf{\implies\:a=\dfrac{10}{3}\:ms^{-2}}}}$

Now, Distance covered by it in 12 sec:-

Again, Using Second Law of Motion:-

$\bf \implies\:S=ut+\dfrac{1}{2}at^2$

Here,

• u = 0

• a = 10/3 m/s²

• t = 12 sec

Putting values,

$\sf \implies\:S=0\times12+\dfrac{1}{2}\times \dfrac{10}{3} \times 12^2$

$\sf \implies\:S=\dfrac{1}{2}\times \dfrac{10}{3} \times 144$

$\sf \implies\:S=\dfrac{10}{3} \times 72$

$\sf \implies\:S=10\times24$

${\boxed{{\bf{\implies\:S=240\:m}}}}$

Hence, The Distance covered by the given body in 12 sec is 240m.

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