Question

An unloaded car moving with velocity u on a frictionless road can be stopped in distance s. If passengers add 40% to its weight and breaking force remains the same, find the stopping distance at velocities?
(a) 1.4s
(b)$\sqrt {1.4} s$
(c) ${1.4^2}s$
(d)$\dfrac{1}{{1.4}}s$

Answer 1

Work done to stop the car is equal to the kinetic energy of the car.

Formula Used:
Kinetic energy:
$K.E. = \dfrac{1}{2}m{v^2}$ ……(1)
Where,
m is mass of object
v is velocity of object
Work done:
$W = \overrightarrow {F \cdot } \overrightarrow S$ ……(2)
Where,
F is force applied
S is distance covered in the direction of force

Step-by-step answer:

Given:
1. Velocity of moving car =u
2. Distance in which unloaded car can be stopped= s
3. Percentage weight added to car= 40%

To find: Distance in which loaded car can be stopped.

Step 1 of 6:
Initially, the car is moving is unloaded and moving with velocity u. Let its mass be m. So, its kinetic energy (K.E.(1)) will be given by eq (1):
$K.E.(1) = \dfrac{1}{2}m{u^2}$ ……(3)
Braking force f is applied to stop the unloaded car and it stops in s distance. Work done (W1) will be given by eq (2):
$W(1) = ( - f)s$ ……(4)

Step 2 of 6:
As kinetic energy is reduced to zero by braking force:
$K.E.(1) = W(1)$
Using eq (3) and (4):
$\dfrac{1}{2}m{u^2} = - fs$ ……(5)

Step 3 of 6:
When passengers are added, the car is moving is still moving with velocity u. Its mass has changed by 40%. Let new mass be m’:
$\begin{gathered} m' = (1 + \dfrac{{40}}{{100}})m \\\ m' = 1.4m \\\ \end{gathered}$

Step 4 of 6:
In case of loaded car, kinetic energy (K.E.(2)) will be given by eq (1):
$K.E.(2) = \dfrac{1}{2}m{u^2}$ ……(6)
Braking force f is applied to stop the loaded car, let’s say it stops in s’ distance. Work done (W2) will be given by eq (2):
$W(2) = ( - f)s'$ ……(7)

Step 5 of 6:
Again, as kinetic energy is reduced to zero by braking force:
$K.E.(2) = W(2)$
Using eq (6) and (7):
$\dfrac{1}{2} \times (1.4m){u^2} = - fs'$ ……(8)

Step 6 of 6:
Comparing eq (5) and (8):
$s' = 1.4s$

Correct Answer:
Distance in which loaded car can be stopped is: (a) 1.4s

Note: In questions like these where 2 cases are given, solve the two separately. Then after you have found the final expressions for both, compare them to get the answer.