JEE Main 2026 — Apr 2, Shift 1 — Physics solution

JEE Main2026Apr 2, Shift 1Physics

A particle is rotating in a circular path and at any instant its motion can be described as $θ = \frac{5 t ^{4}}{40} - \frac{t ^{3}}{3}$ . The angular acceleration of the particle after $10$ seconds is _________ rad/s $^{2}$ .

A
$150$
B
$120$
C
$130$
D
$170$

Solution

Differentiate $θ$ twice with respect to $t$ :

$ω = \frac{d θ}{d t} = \frac{t ^{3}}{2} - t^{2}$

$α = \frac{d ω}{d t} = \frac{3 t ^{2}}{2} - 2 t$

At $t = 10$ s: $α = \frac{3 ( 100 )}{2} - 20 = 150 - 20 = 130$ rad/s $^{2}$ .

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