Question

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Answer 1

★ Provided Question ★

Your school bus accelerate uniformly from 5m/s to 10m/s within 10s of time. Calculate the acceleration produced and distance covered by the bus in that time.

How to solve?

Here, in this question we are already given that its initial velocity(u) is 5m/s and its final velocity (v) is 10m/s.Time taken by it is 10s.

• To calculate the acceleration we will use here the first equation of motion || v = u + at ||.
• To calculate the distance travelled we will use here the second equation of motion || s = ut + $\sf { \dfrac{1}{2}a{t}^{2} }$

So, let's begin to solve!!

★ Required Solution ★

${\underline {\underline {\rm { Given: } }}}$

• Initial velocity (u) = 5m/s
• Final velocity (v) = 10m/s
• Time taken (t) = 10s

${\underline {\underline {\rm { To \: calculate: } }}}$

• Acceleration (a)
• Distance travelled (s)

${\underline {\underline {\rm { Calculation : } }}}$

Calculating acceleration:

We know that:

• ${\boxed {\huge {\bf {\pink { v=u+at}}}}}$

Substituting values:

$\sf { :\implies 10 = 5 + a \times 10 }$

$\sf { :\implies 10 = 5 + 10a }$

$\sf { :\implies 10a = 10 -5 }$

$\sf { :\implies 10a = 5 }$

$\sf { :\implies a = \dfrac{5}{10} }$

$\sf\red { :\implies a = 0.5m/{s}^{2} }$

Therefore, acceleration produced by the bus is $\sf { 0.5m/{s}^{2} }$.

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Calculating distance travelled:

We know that:

• ${\boxed {\huge {\bf {\pink { s =ut + \dfrac{1}{2}a{t}^{2}}}}}}$

Substituting values:

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

$\sf { :\implies s = 50 + \dfrac{1}{\cancel{2}} \times \dfrac{5}{10} \times \cancel{100} }$

$\sf { :\implies s = 50 + \dfrac{5}{\cancel{10}} \times \cancel{50} }$

$\sf { :\implies s = 50 + 5 \times 5 }$

$\sf { :\implies s = 50 + 25}$

$\sf \red{ :\implies s = 75m}$

Therefore distance travelled by the bus in that time is 75m.

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Points to remember:

• When the body starts from the rest then its initial velocity (u) is 0m/s.

• When the body come to stop or applies brake then its final velocity (v) is 0m/s.

Three equations of motion:

• ${\boxed {\huge {\bf {\pink { v=u+at}}}}}$
• ${\boxed {\huge {\bf {\pink { s =ut + \dfrac{1}{2}a{t}^{2}}}}}}$
• ${\boxed {\huge {\bf {\pink { 2as ={v}^{2}-{u}^{2}}}}}}$

Where,

v denotes final velocity.

u denotes initial velocity.

'a' denotes acceleration.

's' or 'h' denotes distance/displacement.

t denotes time.

Laws of motion:

First law of motion:

✤ An object in motion will remain in motion unless acted upon by an outside force. An object at rest will remain at rest unless acted upon by an outside force.

Second law of motion:

✤ The acceleration of the body is directly proportional to the force acting on it and is inversely proportional to its mass and takes place in the same direction of the force.

Third law of motion:

✤For every action there is an equal opposite reaction.

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