Q.A body with mass 5 kg is acted upon by a force F=(−3i^+4j^) N. If its initial velocity at t=0 is v=(6i^−12j^) m s−1, the time at which it will just have a velocity along the y-axis is
Imagine you're tracking a drone flying in the sky. At any instant, it has a position — say, 30 metres east and 40 metres north of you. That's a vector: r=30i^+40j^. A second later, it's moved. The question kinematics asks is: how fast is that position changing? That rate of change is velocity, and to get it, you differentiate the position vector.
But here's the key difference from school calculus: in school, you differentiated a scalar function like y=x2. Here, you're differentiating a vector function — something that has both magnitude and direction, and both can change with time.
The Intuition First
Think of a vector as an arrow. When time passes, that arrow can do two things:
It can get longer or shorter (magnitude changes).
It can rotate (direction changes).
Velocity is the total rate of change of that arrow. If the drone flies straight away from you, only the length changes. If it flies in a circle around you, only the direction changes. Most real motion does both.
So vector differentiation is just: take the derivative of each component separately, because components are independent scalars.
The Precise Statement
If a position vector is written in Cartesian coordinates as:
r(t)=x(t)i^+y(t)j^+z(t)k^
where i^,j^,k^ are fixed unit vectors (they don't change direction with time), then:
dtdr=dtdxi^+dtdyj^+dtdzk^
That's it. You differentiate each component function x(t),y(t),z(t) exactly as you would in single-variable calculus, and the unit vectors stay put.
dtd(f(t)u^)=dtdfu^(if u^ is constant)
Why This Works
The derivative of a vector is defined the same way as for a scalar — as a limit:
dtdr=limΔt→0Δtr(t+Δt)−r(t)
The numerator is a vector difference. When you write r in components, the difference splits into component differences. The limit then acts on each component separately because the unit vectors are constant. So the definition forces component-wise differentiation.
A Concrete Example
A particle moves such that:
r(t)=(3t2)i^+(5sint)j^+(2e−t)k^
Its velocity is:
v(t)=dtdr=(6t)i^+(5cost)j^+(−2e−t)k^
Notice: the x-component grows linearly, the y-component oscillates, the z-component decays. Each derivative is just the ordinary derivative of that component's function.
The One Trap: Non-Constant Unit Vectors
The rule above assumes i^,j^,k^ are fixed. That's true in Cartesian coordinates. But in polar coordinates, the unit vectors r^ and θ^rotate as the particle moves. Differentiating a vector in polar coordinates requires the product rule because the unit vectors themselves depend on time. …
The key idea is that the velocity component along the x-axis must become zero for the velocity to be purely along the y-axis. Using Newton’s second law, we find the acceleration and then apply the kinematic equation for constant acceleration.
The key idea is that the force causes constant acceleration, which changes the x-component of velocity linearly with time. The velocity becomes purely along the y-axis when the x-component becomes zero. Solving vx(t)=0 gives t=10 s.
This is a straightforward application of Newton’s second law and kinematics in vector form. The force is constant, so the acceleration is constant. That means we can treat each component of motion independently — the x and y motions are completely decoupled. The velocity will be along the y-axis when its x-component vanishes, regardless of what the y-component is doing.
Let’s work it through.
Find the acceleration vector.
Newton’s second law: F=ma.
With m=5 kg and F=(−3i^+4j^) N,
a=mF=(−53i^+54j^) m/s2.
Write the velocity as a function of time.
For constant acceleration, v(t)=v0+at.
Initial velocity v0=6i^−12j^ m/s.
So
v(t)=(6i^−12j^)+(−53i^+54j^)t.
Group components:
vx(t)=6−53t,vy(t)=−12+54t.
Set the condition for velocity along the y-axis.
“Just have a velocity along the y-axis” means the x-component is zero (the velocity is purely vertical).
Set vx(t)=0:
6−53t=0⇒t=36×5=10 s.
At t=10 s, vy=−12+54(10)=−12+8=−4 m/s, so the velocity is indeed along the negative y-axis. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
COMEDK 2026Set 2026-A1 markMCQ
Q.A particle moves along a parabolic path y=9x2 in such a way that the x component of velocity remains constant. If, the acceleration of the particle is 2jms−2, find the x component of velocity.
(A) 91ms−1
(B) 41ms−1
(C) 31ms−1
(D) 61ms−1
›Reveal solutionSolution
The key idea is that constant horizontal velocity means zero horizontal acceleration, so the given acceleration is purely vertical. Using the path constraint y=9x2 and differentiating twice with respect to time gives a relation between vx and the vertical acceleration, yielding vx=31m/s.
We are told the particle moves on the parabola y=9x2, and its x-component of velocity is constant. That means vx=x˙ is constant, so x¨=0. The acceleration vector is given as 2j^m/s2, meaning the only acceleration is vertical. This is a classic constrained-motion problem: the path geometry links x and y, so we can differentiate that relation to connect velocities and accelerations.
Write the path equation and differentiate once
The path is y=9x2. Differentiate with respect to time t:
dtdy=9⋅2xdtdx=18xvx.
So the vertical velocity is vy=18xvx.
Differentiate again to get acceleration
Differentiate vy=18xvx with respect to time. Since vx is constant, we treat it as a constant factor:
ay=dtd(18xvx)=18vxdtdx=18vx⋅vx=18vx2.
Notice: we used dtdx=vx and vx constant, so no product rule term from vx itself.
Equate to given acceleration
The problem states the acceleration is 2j^m/s2, so ay=2. Thus:
Q.A body is moving along a straight line with initial velocity v0. Its acceleration a is constant. After t seconds, its velocity becomes v. The average velocity of the body over the given time interval is
(A) vˉ=2atv2+v02
(B) vˉ=atv2+v02
(C) vˉ=2atv2−v02
(D) vˉ=atv2−v02
›Reveal solutionSolution
Average velocity is displacement over time; use v2−v02=2as to express the displacement.
Step 1 — Definition.
vˉ=timedisplacement=ts.
Step 2 — Get s from a kinematic equation.
For uniform acceleration the time-free equation is
v2=v02+2as⟹s=2av2−v02.
We use this particular equation because the options are written in terms of v2 and v02.
Step 3 — Divide by t.
vˉ=ts=2atv2−v02.
Step 4 — Cross-check with the standard result.
For constant acceleration the velocity–time graph is a straight line, so
Q.An object with mass 5 kg is acted upon by a force, F=(−3i^+4j^) N. If its initial velocity at t = 0 is v=(6i^−12j^) ms−1, the time at which it will just have a velocity along y-axis is
(A) 5 s
(B) 10 s
(C) 2 s
(D) 15 s
›Reveal solutionSolution
The key idea is that the velocity component along the x-axis must become zero for the velocity to be purely along the y-axis. Using Newton’s second law and kinematic equations, the time comes out to be 10 s.
The problem asks for the time when the velocity is only along the y-axis. That means the x-component of velocity must be zero at that instant. The force is constant, so the acceleration is constant — this is a straightforward application of the equations of motion under uniform acceleration.
Let’s break it down.
Find the acceleration vector.
Newton’s second law: F=ma.
Given m=5 kg and F=(−3i^+4j^) N,
a=mF=(5−3i^+54j^) m/s2.
Write the velocity as a function of time.
For constant acceleration,
v(t)=u+at,
where initial velocity u=(6i^−12j^) m/s.
So
v(t)=(6−53t)i^+(−12+54t)j^.
Set the x-component to zero.
For velocity to be purely along the y-axis, the i^ component must vanish:
6−53t=0.
Solve:
53t=6⇒t=6×35=10 s.
Check the y-component at that time.
At t=10 s,
vy=−12+54(10)=−12+8=−4 m/s. …