Question

A car of mass $1200kg$ is moving at $90km/h$ when the driver observes the need to apply the brakes. The braking force is increasing linearly with time at the end of $3600N/s$. Find the time in which the car comes to halt?

Answer 1

Form the linear equation of force in terms of $t$. Then write force in terms of acceleration. And using acceleration, find the relation between velocity and time to solve the question.

Complete step by step answer: It is given in the question that.
$m = 1200kg$ is the mass of the car
$u = 90cm/h$ is the initial velocity of car
$\Rightarrow u = 90 \times \dfrac{{1000}}{{60 \times 60}}m/s$
$\Rightarrow u = 25m/s$
Now, it is given that the breaking force is increasing linearly with respect to time to $3600N/s$.
$\Rightarrow \dfrac{f}{t} = - 3600$ . . . (1)
Where,
$f$ is the breaking force
$t$ is time
Now, we know that
$F = ma$
Where,
$F$ is force applied
$m$ is mass of the body
$a$ is the acceleration
Therefore, by substituting the given values in the above equation. We get
$f = ma$
$\Rightarrow - 3600t = 1200a$
By rearranging it, we get
$\dfrac{{ - 3600t}}{{1200}} = a$
$\Rightarrow a = - 3t$
By the definition of acceleration. Acceleration is the rate of change of velocity with respect to time.
$\Rightarrow a = \dfrac{{dv}}{{dt}}$
$\Rightarrow \dfrac{{dv}}{{dt}} = - 3t$
By rearranging it, we get
$dv = - 3tdt$
When $t = 0$, the initial velocity is $u$
When $t = t$, the final velocity is $v$
By integrating the above equation in these limits, we can write
$\int\limits_u^v {dv = \int\limits_0^t { - 3tdt} }$
$\Rightarrow \left[ v \right]_u^v = - 3\left[ {\dfrac{{{t^2}}}{2}} \right]_0^t$ $\left( {\because \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right)$
By substituting the upper and lower limits, we get
$v - u = - \dfrac{3}{2}\left( {{t^2} - 0} \right)$
$\Rightarrow v - u = \dfrac{{ - 3{t^2}}}{2}$
It is given that,
$u = 25m/s$
Therefore, we get
$v = \dfrac{{ - 3{t^2}}}{2} + 25$ . . . (1)
Since, the car stops after breaking. The final velocity of the car will be equal to zero.
i.e. $v = 0$
Therefore, equation (1) becomes
$0 = \dfrac{{ - 3{t^2}}}{2} + 25$
By rearranging we get,
$\Rightarrow \dfrac{{3{t^2}}}{2} = 25$
By cross multiplying we, get
$\Rightarrow 3{t^2} = 50$
Dividing both the sides by 3, we get
$\Rightarrow {t^2} = \dfrac{{50}}{3}$
$\Rightarrow t = \sqrt {\dfrac{{50}}{3}} s$
Therefore, the time in which the car comes to halt is $\sqrt {\dfrac{{50}}{3}} s$

Note: The force applied for breaking is written with a negative sign. It is because the force is applied in the opposite direction of the motion of the car. That is why the acceleration is also negative, because negative acceleration brings the car to halt and the positive acceleration increases the speed of the car.