Question

A body moves in a straight line with a
uniform arceration. If the initial velocity
of the body is 5m/s
after 10 second is 15 m/s find the
acceleration and distance covered by
the body in this time interval​

Answer 1

Answer :

➥ Acceleration of the body = 1 m/s²

➥ Distance coverd by the body = 100 m

Given :

➤ Intial velocity of the body (u) = 5 m/s

➤ Final velocity of the body (v) = 15 m/s

➤ Time taken by the body (t) = 10 s

To Find :

➤ Acceleration of the body (a) = ?

➤ Distance coverd by the body (s) = ?

Solution :

As we know that

$\tt{ : \implies a = \dfrac{v - u}{t} }$

$\tt{:\implies a = \dfrac{15 - 5}{10} }$

$\tt{:\implies a = \cancel{ \dfrac{10}{10}}}$

$\tt{:\implies \green{\underline{ \overline{ \boxed{ \purple{ \bf{ \: \: a = 1 \: m/ {s}^{2} \: \: }}}}}}}$

Hence, the acceleration of the body is 1 m/s².

Now ,

We calculate distance covered by the body

From second equation of motion

$\tt{:\implies s = ut + \dfrac{1}{2} a{t}^{2} }$

$\tt{:\implies s = 5 \times 10 + \dfrac{1}{2} \times 1 \times {10}^{2} }$

$\tt{:\implies s = 50 + \dfrac{1}{2} \times 1 \times 100 }$

$\tt{:\implies s = 50 + \dfrac{1}{ \cancel{ \: 2 \: }} \times \cancel{100}}$

$\tt{:\implies s = 50 + 50}$

$\tt{:\implies \green{\underline{ \overline{ \boxed{ \purple{ \bf{ \: \: s = 100 \: m\: \: }}}}}}}$

Hence, the distance coverd by the body is 100 m.

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Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as

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