A car of mass 1000 kg is moving with a certain speed when a constant braking force 1000 N acts on it for 5 s and the speed of the car reduces to half of its original speed. Find the further time required to stop the car if the same constant force acts.


Given,
Mass of car, m = 1000 kg
Force acting, F = 1000 N
Time taken, t = 5 s
Initial speed, u = u (supposed)
Final speed, v = u/2 $[ \because initial\ speed\ is\ reduced\ to\ half]$

To find
t = ?, when v = 0

Solution:
We are dividing the question into two-part.

First part, when a car of mass 1000 kg is moving with a certain speed and after applying a constant braking force of 1000 N for 5 s then the speed of the car reduces to half of its original speed. 

Here, as the brake is applied the speed reduces, so the car is said to be retarded means negatively accelerated (-a).

By Newton's second law of motion, magnitude of force is gievn as-
$F=m\times a$
$Retardation(a)=\frac{F}{m}$
$a=\frac{1000N}{1000kg}$
$a=1m/{s}^{2}$
$a=-1m/{s}^{2}$ $[ \because negative\ acceleration\ or\ retardation]$

To get initial speed u from known retardation a, we'll use the equation of motion.  
$v=u+(a\times t)$    
Then, $u=v-a\times t$     ------------------- (i)
Substituting the given value in (i)
$u=\frac{u}{2}-[(-1)\times 5]$
$u=\frac{u}{2}+5$
$u=\frac{u+10}{2}$
$2u=u+10$
$u=10m/s$

Therefore, the initial speed of the car must have been 10 m/s.
And, the final speed of the car will be 5 m/s. (the speed of the car reduces to half of its initial speed)
 
Now, the Second part of the question is how much further time will be required to stop the car if the same constant force acts.
Here, 
Initial speed, u = 5 [after applying breaks the final speed now acts as the initial speed when the same constant force is applied from this point to stop the car]
Final velocity, v = 0  [as car stops]
Acceleration, a = -1 [ifthe same constant force is applied, then acceleration will be the same]
Time, t = ?

To get time t, we will use the same equation of motion.
Substituting the given value of second part of the question in (i)
$5=0-(-1)\times t$
$5=1\times t$
$t=5sec$

Hence, the time required to completely stop the car is $5sec$.

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Updated on: 10-Oct-2022

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