Question

The velocity of a car increases from 30 kmh-1 to 60 kmh-1 in 30 seconds. Calculate the acceleration of the car and distance travelled by the car in these 30 s.



answer it ​

Answer 1

$\large \dag\ \underline{\mathbb{GIVEN :-}}$

$\textsf{The velocity of a car increases from 30 km/h to 60 km/h}\\\textsf{ in 30 seconds.}$

$\bullet\ \textsf{Initial velocity(u) = 30 km/h}$

$\bullet\ \textsf{Final velocity(v) = 60 km/h}$

$\bullet\ \textsf{Time(t) = 30 secs}$

$\large \dag\ \underline{\mathbb{TO\ FIND :-}}$

$\textsf{The acceleration(a) of the car and the distance travelled(s)}\\\textsf{ by it.}$

$\large \dag\ \underline{\mathbb{FORMULAS\ TO\ BE\ USED :-}}$

$\bigstar\ \sf{a=\dfrac{v-u}{t} }$

$\bigstar\ \sf{v^2-u^2=2as}$

$\sf{\longmapsto a\ is\ the\ acceleration.}\\\sf{\longmapsto u\ is\ the\ initial\ velocity.}\\\sf{\longmapsto v\ is\ the\ final\ velocity.}\\\sf{\longmapsto t\ is\ the\ time\ taken.}\\\sf{\longmapsto s\ is\ the\ distance\ covered.}$

$\large \dag\ \underline{\mathbb{SOLUTION :-}}$

$\sf{u=30\ km/h=30\times \dfrac{5}{18}=\dfrac{25}{3}=8.33\ m/s }$

$\sf{v=60\ km/h=60\times \dfrac{5}{18} =\dfrac{50}{3}=16.67\ m/s }$

$\textsl{Putting the given values :-}$

$\sf{a=\dfrac{16.67-8.33}{30} }\\\\\sf{\implies a=\dfrac{8.34}{30} }\\\\\boxed{\sf{\implies a=0.278\ m/s^2 }}$

$\rule{150}2$

$\textsl{Again putting the values :-}$

$\sf{(16.67)^2-(8.33)^2=2\times 0.278 \times s}\\\\\sf{\implies 277.8889-69.3889=0.556s}\\\\\sf{\implies 208.5=0.556s}\\\\ \sf{\implies s=\dfrac{208.5}{0.556} }\\\\\boxed{\sf{\implies s=375\ m}}$

$\underline{\textbf{Hence, the acceleration of the car is 0.278 m/s}^2 \mathbf{and}}\\\underline{\textbf{the distance covered by it is 375 m.}}$

Answer 2

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$\huge{ \sf{ \underline{ \underline{ \purple{Answer...}}}}}$

♣️ GiveN:

• Velocity increases from 30 kmh^-1 to 60 kmh^-1

♣️ To FinD:

• Accleration of the car in 30 s
• Distance travelled in 30 s

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$\huge{ \underline{ \underline{ \sf{ \purple{Explanation...}}}}}$

The above question is very simple, just apply formula and get the solution type question. But we have to use the equations of motion to solve this. So, we need to know about the three equations of motion that are:

$\large{ \sf{ \red{v = u + at.........(1)}}} \\ \\ \large{ \sf{ \red{s = ut + \frac{1}{2}a {t}^{2} ........(2)}}} \\ \\ \large{ \sf{ \red{ {v}^{2} - {u}^{2} = 2as.........(3)}}}$

✍ Note:

• Symbols have their usual meanings.

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So, applying these formulae, we gonna solve the Q.

But before that, we have to convert the units because of disimilar units.

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$\huge{ \sf{ \underline{ \underline{ \purple{Unit \: conversion...}}}}}$

Given, time taken = 30 s

Initial velocity = 30 km/h

Converting into m/s,

$\large{ \sf{30 \: kmh {}^{ - 1} = \frac{30000 \: m}{3600 \: s} = \frac{25}{3} m {s}^{ - 1} }}$

Final velocity = 60 km/h

Converting into m/s,

$\large{ \sf{60 \: km {h}^{ - 1} = \frac{60000 \: m}{3600 \: s} = \frac{50}{3}m {s}^{ - 1} }}$

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$\huge{ \underline{ \underline{ \purple{ \sf{Solution...}}}}}$

We have been given, the initial velocity, final velocity and time taken, and it ask to find accleration.

By using 1st equation of motion

$\large{ \sf{ \mapsto \: v = u + at}} \\ \\ \large{ \sf{ \mapsto \: \frac{50}{3}m {s}^{ - 1} = \frac{25}{3}m {s}^{ - 1} + a \times 30s}} \\ \\ \large{ \sf{ \mapsto \: \frac{50}{3} - \frac{25}{3} = a \times 30}} \\ \\ \large{ \sf{ \mapsto \: 30a = \frac{25}{3}}} \\ \\ \large{ \sf{ \mapsto \: a = \cancel{ \frac{25}{3 \times 30}\: m {s}^{ - 2}}}} \\ \\ \large{ \sf{ \mapsto \: a = \frac{5}{18} m {s}^{-2}}}\\ \\ \large{ \sf{ \mapsto \: \boxed{ \sf{ \red{a = 0.2 \bar{7} \: m {s}^{ - 2} }}}}} \\ \\ \large{ \therefore{ \sf{ \underline{ \green{accleration \: of \: the \: body = 0.2 \bar{7}m {s}^{ - 2}}}}}}$

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Now, we have the accleration, initial and final velocity and also time, we have to find the distance.

So, w can use anyone of the 2nd and 3rd equations of motions, I am using the 3rd equation, it's your choice.

By using 3rd equation of motion,

$\large{ \sf{ \mapsto \: {v}^{2} - {u}^{2} = 2as}} \\ \\ \large{ \sf{ \mapsto \: {( \frac{50}{3}) }^{2} - {( \frac{25}{3}) }^{2} = 2 \times \frac{5}{18} \times s}} \\ \\ \large{ \sf{ \mapsto \: \frac{2500}{9} - \frac{625}{9} = \frac{5s}{9} }} \\ \\ \large{ \sf{ \mapsto \: \frac{5s}{ \cancel{9}} = \frac{1875}{ \cancel{9}} }} \\ \\ \large{ \sf{ \mapsto \: s = \frac{1875 \: m}{5}}} \\ \\ \large{ \sf{ \mapsto \: \boxed{ \sf{ \red{s = 375 \: m}}}}} \\ \\ \large{ \therefore{ \underline{ \sf{ \green{distance \: travelled \: in \: 30 s = 375 \: m}}}}}$

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