Question

if a car accelerates uniformly from 18 km/hr to 36 km/hr in 5 seconds.

Answer 1

Given :-

If a car accelerates uniformly from 18 km/hr to 36 km/hr in 5 seconds

To Find :-

Acceleration

Solution :-

1 km/h = 5/18 m/s

18 km/h = 18 × 5/18

18 km/h = 5 m/s

36 km/h = 36 × 5/18

36 km/h = 2 × 5

36 km/h = 10 m/s

Now,

Acceleration = Final velocity - Initial velocity/Time

a = 10 - 5/5

a = 5/5

a = 1/1

a = 1

Therefore

Acceleration is 1 m/s²

Answer 2

☆Given that

• Initial velocity = u = 18 km/m
• Final velocity = v = 36 km/m
• Time taken = t = 5 seconds

☆To find

• Accelerates of car
• Distance travelled

☆Solution

Given Parameters : - Time taken by car be 5 sec and Initial velocity = 18km/hours &Final velocity = 36km/hours. we have to find Accelerates

converting and final velocity in m/s

$\underline{ \bf \red{ \: conversion \: from \: k m/s \: velocity}}$

$\tt\purple \longrightarrow 1km = 1000 \\ \\ \\ \tt\purple \longrightarrow 1 \: hours \: = 60 \: min \: = 3600 \: sec$

$\tt\purple \therefore \: \: \dfrac{1km}{hours} = \dfrac{10 \cancel{00 }\:m}{36 \cancel{00} \: s} \\ \\ \\ \tt\purple \therefore \: \: \: \frac{1km}{hours} = \cancel \frac{10}{36} \\ \\ \\ \tt\purple \therefore \: \: \: 1km/h = \: \frac{5}{18}$

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$\bf Initial \: velocity _{(u)} = 18km/hours \\ \\ \\ \red\longrightarrow \cancel{18} \times \frac{5}{ \cancel{18}} \\ \\ \\ \red\longrightarrow5 \sf \: m/s$

$\bf Final \: velocity _{(v)}=36 km/hours \\ \\ \\ \red\longrightarrow \cancel{36 }\times \frac{5}{ \cancel{18} } \\ \\ \\ \red\longrightarrow \cancel6 \times \frac{5}{ \cancel3} \\ \\ \\ \red\longrightarrow2 \times 5 \\ \\ \\ \red\longrightarrow10 \sf \: m/s$

____________________________

we need to calculate Acceleration and distance traveled

we know that

$\qquad \qquad\boxed{\underline{\frak{ Acceleration_{(a) }= \frac{(v - u)}{t} }}}$

where

• v = Final velocity
• u = Initial velocity
• t = time

Substituting the given values in the above equation we get

$\sf \: \blue\longrightarrow \: a \: = \dfrac{(10 - 5)}{5} \\ \\ \\ \blue\longrightarrow \sf \: a = \cancel\dfrac{ 5}{5} \\ \\ \\ \blue\longrightarrow \sf \: Acceleration = 1m/s²$

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we know that distance traveled is calculated by the formula

$\qquad \qquad\boxed{\underline{\frak{Distance_{ (travelled )}=ut + \bigg ( \frac{1}{2} \bigg )a \times t^2}}}$

Substituting the given value in the above equation we get

$\tt\blue\longrightarrow \: s = 5 \times 5 + \bigg( \dfrac{1}{2} \bigg) \times 1 \times {(5)}^{2} \\ \\ \\ \tt\blue\longrightarrow \: s = 25 + \bigg( \dfrac{1}{2} \bigg) \times 5 \times 5 \\ \\ \\ \tt\blue\longrightarrow \: s = 25 + \dfrac{1}{ \cancel2} \times \cancel{25} \\ \\ \\ \tt\blue\longrightarrow \: s = 25 + 12.5 \\ \\ \\ \tt\blue\longrightarrow \: s = 37.5 \: m$

Hence Acceleration = 1 m/s² and Distance travelled = s = 37.5 m