VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Equation of motions is considered one the most difficult concept.
83 Questions around this concept.
A particle has an initial velocity and an acceleration of
Its speed after 10s is
An engine of the train, moving with uniform acceleration, passes the signal -post with velocity u and the last compartment with velocity v. The velocity with which the middle point of the train passes the signal post is:
Speeds of two identical cars are and
at a specific instant. If the same deceleration is applied on both cars, the ratio of the respective distances in which the two cars are stopped from that instant is :
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A car moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is :
An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, 120 km/h, the stopping distance will be :
A particle is moving with velocity is a constant. The general equation for its path is
A train is standing on a platform, and a man inside a compartment of a train drops a stone. At the same instant train starts to move with constant acceleration. The path of the particle as seen by the person who drops the stone is :
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
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A car travelling at a speed of 60 km/hr can break and stop within a distance of 20 metres. If the car is travelling at twice the speed, i.e.,120 km/hr the stopping distance will be:
Two balls A and B are thrown from a building, A is thrown up and B is thrown down (both vertically). Let and
be their respective speeds when they reach the ground, then:
The acceleration of a particle increases with time “t” as From the initial position with a velocity u; the distance travelled by the particle in time “t” is :
There are three equations of motion
The first kinematical equation of motion (velocity-time equation)
Formula
$
v=u+a t
$
$v=$ Final velocity
$\mathrm{u}=$ Initial velocity
$A=$ acceleration
$
\mathrm{T}=\text { time }
$
2. The second kinematical equation of motion (Position-time equation)
Formula
$s=u t+\frac{1}{2} a t^2$
$s \rightarrow$ Displacement
$u \rightarrow$ Initial velocity
$a \rightarrow$ acceleration
$t \rightarrow$ time
3. The third kinematical equation of motion (Velocity-displacement equation)
Formula
$
v^2-u^2=2 a s
$
$v \rightarrow $ Final Velocity
$s \rightarrow$ Displacement
$u \rightarrow$ Initial velocity
$a \rightarrow$ acceleration
Displacement in nth second
Formula: $S_n=u+\frac{a}{2}(2 n-1)$
Where $u=$ Initial velocity
$a=$ uniform acceleration
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