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Physics Question 84 – AP-EAMCET 2025

For a particle moving in x-y plane, if at any instant of time 't', (in second) its displacements (in metre) are x=2t2−t and y=4t2−4t, then the velocity of the particle at a time t=1 s is

Velocity is the rate of change of displacement with respect to time. For motion in a plane, the velocity can be found by differentiating the position vector components.

Step 1: Understand the given displacement equations✦ Active

The position of the particle in the x-y plane is given by its displacement components as functions of time:

x(t)=2t2−t
y(t)=4t2−4t

We need to find the magnitude of the velocity at t=1 s.

Step 2: Determine the velocity components by differentiation○ Expand

Velocity is the time derivative of displacement. We differentiate x(t) and y(t) with respect to time t to find the velocity components vx(t) and vy(t).

vx=dxdt=ddt(2t2−t)=4t−1
vy=dydt=ddt(4t2−4t)=8t−4
💡 Teacher's Secret Hint

Remember the power rule for differentiation: ddt(atn)=natn−1 and ddt(ct)=c.

Step 3: Calculate the velocity components at the specified time○ Expand

Substitute t=1 s into the expressions for vx(t) and vy(t):

vx(t=1 s)=4(1)−1=3 m/s
vy(t=1 s)=8(1)−4=4 m/s
💡 Teacher's Secret Hint

Ensure units are consistent throughout the calculation. Here, displacement is in meters and time in seconds, so velocity will be in m/s.

Step 4: Calculate the magnitude of the velocity○ Expand

The magnitude of the velocity vector v→=vxi^+vyj^ is given by v=vx2+vy2.

v=(3 m/s)2+(4 m/s)2
v=9+16
v=25
v=5 m/s

The velocity of the particle at t=1 s is 5 m s−1.

💡 Teacher's Secret Hint

This is a classic 3-4-5 Pythagorean triplet, which often appears in physics problems involving vector magnitudes.

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