Question

A car moves from rest with uniform acceleration and acquire the velocity of 15m/s in 50 sec. Find the acceleration and distance covered by car in this period​

Answer 1

Explanation:

Distance =speed ×time

=15×50

=750m

Acceleration =velocity/ time

=15/50

=0.3m/s

Hope you have gote your answer

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,

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

Here,

u = 0

S = 15m

t = 3 sec

Putting the values,

$\sf \implies\:15=0\times3+\dfrac{1}$

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

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

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

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

Now,

Distance covered by it in 12 sec:-

Again,

Using Second Law of Motion:-

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

Here,

u = 0

a = 10/3 m/s²

t = 12 sec

Putting values,

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

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

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

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

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

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

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