Question

If the bike's velocity = 72 km/hr. I stop the bike with counter acceleration of 8 $m/{{s}^{2}}$. So how much time does the bike take to become motionless.

Answer 1

In this question we are asked to calculate the time required for the bike to come at rest when initially it is moving with given velocity of 72 km/hr. To solve this question, we will use the laws of kinematics. We need to find the relation between the acceleration, velocity and time. Therefore, we will be using the second law of kinematic motion.

Formula used:
$v=u+at$
Where,
v is the final velocity
u is the initial velocity
a is the acceleration of body
t is the time required

Complete answer:
We have been given that counter acceleration is applied to the bike moving with initial velocity of 72 km/hr. Therefore, the time required for the bike to come to rest will be given by equation,
$v=u+at$
The final velocity of the bike will be taken as zero as the bike will come to rest after some time. Also, the initial velocity needs to be converted into m/s.
Therefore,
We get,
$72km/hr=20m/s$
Therefore, after substituting the given values in above equation
We get,
$0=20-8t$ ………………….. (since acceleration is in opposite direction)
Therefore,
$t=\dfrac{20}{8}$
We get,
$t=2.5\sec$
Therefore, the bike moving with velocity of 72 km/hr, when applied counter acceleration of 8 $m/{{s}^{2}}$will come to rest in 2.5 seconds.

Note:
The kinematic equations can be paired in three ways: position – time, velocity - time, velocity – position. Therefore, these laws are a set of equations that describe the motion of an object with constant acceleration. These laws can not be used for an object that has varying acceleration. Using the kinematic equations we can calculate the unknown parameter of the motion of the body, if two of the three parameters are given.