A particle moves along the x-axis. Its position as a function of time is given by: x(t) = 3t³ - 9t² + 5 Find the time at which the particle momentarily comes to rest and the acceleration at that instant.

Velocity: v = dx dt = 9t² - 18t Set v = 0: 9t(t - 2) = 0 → t = 0 or t = 2 s Acceleration: a = dv dt = 18t - 18 At t = 2 s: a = 18(2) - 18 = 18 m/s²

JEE physics

A particle moves along the x-axis — JEE Physics

LLiamAnderson39 Asked 2mo ago 1,091 views 1 answer

A particle moves along the x-axis. Its position as a function of time is given by:

$$x(t) = 3t^3 - 9t^2 + 5$$

Find the time at which the particle momentarily comes to rest and the acceleration at that instant.

TSTushar Saxena ✓ Accepted · 2mo ago ▲ 24

Velocity: $v = \dfrac{dx}{dt} = 9t^2 - 18t$

Set $v = 0$: $9t(t - 2) = 0 \Rightarrow t = 0$ or $t = 2$ s

Acceleration: $a = \dfrac{dv}{dt} = 18t - 18$

At $t = 2$ s: $a = 18(2) - 18 = 18 m/s^2$

For revision — the key formula used here comes up almost every year.

SandeepRanasinghe88 · 2mo ago

Is there a faster shortcut for this in the actual exam? Time is tight.

AnanyaPatel44 · 2mo ago

Why do we take the positive value only in the last step?

Siddharth Das · 2mo ago