Kinematics (JEE Main)

Question 1

A particle starts from rest at x = 0 and moves along the +x axis with velocity v = β√x, where β is a positive constant. Over the interval in which it covers its first s metres, let v̄ₜ be its time-ave

Accepted Answer

C. Key insight: the two averages weight the motion differently, so they need not be equal. From dx/dt = β√x we get 2√x = β t, so the time to cover s is tₛ = 2√s/β. Time average: v̄ₜ = s/tₛ = β√s/2. Space average: v̄ₓ = 1/s∫₀^s β√x dx = 2/3β√s. Ratio = (2/3)/(1/2) = 4/3. Trap: assuming both averages are the same (answer 1), or inverting the ratio (3/4). The space average is larger because the particle

Question 2

A particle moving along a straight line with initial speed v₀ experiences a retardation of magnitude k√v, where v is its instantaneous speed and k is a positive constant. Its average speed over the en

Accepted Answer

C. Key insight: the particle stops in finite time and finite distance, but the average is not v₀/2 because the retardation is not constant. Time: dv/dt = −k√v gives 2√(v₀) = kT, so T = 2√(v₀)/k. Distance: v dv/dx = −k√v gives dx = −√v dv/k, so X = 2/3v₀^(3/2)/k. Average speed = X/T = (2v₀^(3/2)/3k)/(2√(v₀)/k) = (v₀)/3. Trap: v₀/2 is the uniform-retardation result; here the speed falls fastest at the

Question 3

A motorboat moving in a straight line switches off its engine when its speed is 40 m/s. Thereafter the water produces a retardation proportional to the boat's speed. The boat covers 60 m while its spe

Accepted Answer

80. Key insight: with a = −kv, write v dv/dx = −kv, so dv/dx = −k: the speed falls linearly with distance even though it decays exponentially with time. Hence v = v₀ − kx. From the data, 10 = 40 − 60k, so k = 0.5 m⁻¹. The boat stops (v → 0) at x = v₀/k = 40/0.5 = 80 m. Trap: since v = v₀e^(−kt) never becomes zero, students conclude the distance is infinite; the distance actually converges to 80 m. Exa

Kinematics: questions with solutions

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