The position vector of particle changes with time according to the relation r(t) = 15t^2i + (4 - 20t^2)j. What is the magnitude of the acceleration (in ms^-2) at t = 1?

The position vector of particle changes with time according… — JEE Main 2019 Physics PYQ

JEE Main 2019 · 9 Apr · Shift II

The position vector of particle changes with time according to the relation

\vec{r}(t) = 15t^2\hat{i} + (4 - 20t^2)\hat{j}

. What is the magnitude of the acceleration (in ms

^{-2}

) at

t = 1

A.
50
B.
100
C.
25
D.
40

Show answer and solution

Correct option: A

Solution

Differentiate r(t) twice to get acceleration: v(t) = 30t î + (-40t) ĵ, then a = 30 î - 40 ĵ (constant, independent of t). Magnitude = √(30² + 40²) = √(900 + 1600) = √2500 = 50 ms⁻².

[JEE Main 2019 · 9 Apr · Shift II]

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