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Question: A car starting from rest has a speed of \(30\,kmh{r^{ - 1}}\) at any one instant. Two seconds later ...

A car starting from rest has a speed of 30kmhr130\,kmh{r^{ - 1}} at any one instant. Two seconds later its speed is 36kmhr136\,kmh{r^{ - 1}} and two seconds after that it is 42kmhr142\,kmh{r^{ - 1}}.What is its acceleration in ms2m{s^ - }^2?

Explanation

Solution

This problem can be solved using the first equation of motion. First convert the given speeds and calculate acceleration using the equation of motion. The car starts from rest, hence its initial velocity will be 0. After two seconds the velocity is given. Also, the thing to do first is to convert all the given quantities in standard SI units.

Formula Used:
To solve the problem we have used the following formula,
v=u+atv = u + at v=u+atv = u + at
Where vv represents the final velocity, uu represents the initial velocity, aa represents acceleration and tt represents time taken.

Complete step by step answer:
In the question it is given that when the car starts from rest its speed is 30kmhr130kmh{r^{ - 1}} which means its initial velocity is 30kmhr130kmh{r^{ - 1}}. Two seconds later its speed is 36kmhr136kmh{r^{ - 1}}and two seconds after that it is 42kmhr142kmh{r^{ - 1}}. So, we have to first calculate the acceleration in ms2m{s^ - }^2. Firstly, we will have to convert the speeds from kmhr1kmh{r^{ - 1}} toms1m{s^{ - 1}}. So, converting each of the given velocities we can write,

u1=30kmhr1 u1=30×10003600ms1 u1=30036ms1 u1=253ms1{u_1} = 30kmh{r^{ - 1}} \\\ \Rightarrow{u_1} = \dfrac{{30 \times 1000}}{{3600}}m{s^{ - 1}} \\\ \Rightarrow{u_1} = \dfrac{{300}}{{36}}m{s^{ - 1}} \\\ \Rightarrow{u_1} = \dfrac{{25}}{3}m{s^{ - 1}}

u2=36kmhr1 u2=36×10003600ms1 u2=36036ms1 u2=10ms1 \Rightarrow{u_2} = 36kmh{r^{ - 1}} \\\ \Rightarrow{u_2} = \dfrac{{36 \times 1000}}{{3600}}m{s^{ - 1}} \\\ \Rightarrow{u_2} = \dfrac{{360}}{{36}}m{s^{ - 1}} \\\ \Rightarrow{u_2} = 10m{s^{ - 1}}

u3=42kmhr1 u3=42×10003600ms1 u3=42036ms1 u3=353ms1 \Rightarrow{u_3} = 42kmh{r^{ - 1}} \\\ \Rightarrow{u_3} = \dfrac{{42 \times 1000}}{{3600}}m{s^{ - 1}} \\\ \Rightarrow{u_3} = \dfrac{{420}}{{36}}m{s^{ - 1}} \\\ \Rightarrow{u_3} = \dfrac{{35}}{3}m{s^{ - 1}}

Now, for calculating acceleration from the formula we use the values provided in the question as,
u=u1=253ms1u = {u_1} = \dfrac{{25}}{3}m{s^{ - 1}}
v=u2=10ms1\Rightarrow v = {u_2} = 10m{s^{ - 1}}
t=2s\Rightarrow t = 2s........................(given in the question)
Therefore, substituting the respective values of u,v,tu,v,t in the formula we get,
v=u+atv = u + at
10=253+2a\Rightarrow 10 = \dfrac{{25}}{3} + 2a
2a=10253\Rightarrow 2a = 10 - \dfrac{{25}}{3}
2a=30253 2a=53\Rightarrow 2a = \dfrac{{30 - 25}}{3} \\\ \Rightarrow 2a= \dfrac{5}{3}
a=53×2 a=56ms2\Rightarrow a = \dfrac{5}{{3 \times 2}} \\\ \therefore a= \dfrac{5}{6}m{s^ - }^2

Hence, acceleration is 56ms2\dfrac{5}{6}m{s^ - }^2.

Note: This problem requires conversion of units many times. So, the conversions must be done carefully so that all the quantities are in the same system of units. Also, it is important to understand which of the given velocity represents initial velocity and which one represents final velocity. While doing problems involving Newton’s equations of motion, we have to keep in mind the sign conventions. If a body is accelerating then its acceleration is positive and if the body is coming to stop after some time then it is accelerating and its acceleration is negative.