Two cars of same mass are moving with velocities v1 and v2 respectively. If they are stopped by supplying same breaking power in time t1 and t2 respectively then v1/v2 is?
√t1/√t2
To solve this problem, we need to use the concept of work and energy. The work done by the braking power is equal to the change in kinetic energy of the cars.
The work done by the braking power is given by the formula:
Work = force x distance
Since the braking force is constant, we can rewrite the equation as:
Work = force x distance = mass x acceleration x distance
Now, let's consider the first car with velocity v1 and time t1 to stop. The distance covered by the first car, d1, can be calculated using the equation of motion:
d1 = (v1 x t1) - (1/2 x a x t1^2)
Here, 'a' is the deceleration due to braking, which can be calculated using the formula:
a = (v1 - 0) / t1 = v1 / t1
Substituting this value for 'a' into the equation above, we get:
d1 = v1 x t1 - (1/2) x (v1/t1) x t1^2
= v1 x t1 - (1/2) x v1 x t1
= (1/2) x v1 x t1
Similarly, for the second car with velocity v2 and time t2 to stop, the distance covered, d2, is:
d2 = (1/2) x v2 x t2
Now, let's equate the work done by the braking power for both cars:
Work1 = Work2
(mass x acceleration x distance1) = (mass x acceleration x distance2)
Since the masses and accelerations are the same for both cars, we can cancel them out:
distance1 = distance2
(1/2) x v1 x t1 = (1/2) x v2 x t2
Now, we can simplify the equation:
v1 x t1 = v2 x t2
And rearranging the equation, we get:
v1 / v2 = t2 / t1
Therefore, v1/v2 is equal to the ratio of the times taken by the second car to stop to the first car to stop.