Question

Find the displacement of a car which increases its speed from $20m{\text{ }}{{\text{s}}^{ - 1}}$ to $80m{\text{ }}{{\text{s}}^{ - 1}}$ in $12\sec$.

Answer 1

To solve this question, we must have a concept of motion and all the equations of motion given by Newton. Here firstly we find the acceleration of the car by simply substituting the values in the first equation of motion and then after we put that value of acceleration along with other values in equation third and solving, we get our required solution.

Formula used:
$v = u + at$
$\Rightarrow {v^2} = {u^2} + 2as$
Where, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, $t$ is the time taken and $s$ is the displacement.

Complete step by step answer:
According to the question it is given that,
$u = 20\,m{\text{ }}{{\text{s}}^{ - 1}}$
$\Rightarrow v = 80\,m{\text{ }}{{\text{s}}^{ - 1}}$ and
$\Rightarrow t = 12\,\sec$
Now we will simply apply the formula from Newton’s law of motion and will find acceleration.
$v = u + at$
Now substituting all the given values and solving for $a$ .
$80 = 20 + a \times 12 \\\ \Rightarrow a = \dfrac{{60}}{{12}} \\\ \Rightarrow a = 5\,m{\text{ }}{{\text{s}}^{ - 2}} \\\$
Therefore, the acceleration of the body is $5m{\text{ }}{{\text{s}}^{ - 2}}$. Similarly substituting the values in the third equation of motion, i.e., ${v^2} = {u^2} + 2as$ and solving for the displacement,
${v^2} = {u^2} + 2as \\\ \Rightarrow {\left( {80} \right)^2} = {\left( {20} \right)^2} + 2 \times 5 \times s \\\ \Rightarrow s = \dfrac{{6000}}{{10}} \\\ \therefore s = 600\,m \\\$
Hence, the displacement of the car is $600\,m$.

Note: In the above given values all the values are in standard form so we don’t have to convert any of the values and simply solve them. But check this out for another question.This will make you make mistakes so convert all the units in standard form and solve them.And member all the equations of motion to solve this type of question.