The position vector of a particle changes with time according to the relation $\vec{r}(t)=15 t^{2} \hat{i}+(4-20 t^{2}) \hat{j}$. What is the magnitude of the acceleration at $t=1$?

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Kinematics

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

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

. What is the magnitude of the acceleration at

t = 1

(A)40
(B)100
(C)25
(D)50✓ Correct

Answer: (D)

Solution

The acceleration is the second time-derivative of position. Differentiating

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

once gives

v = 30 t \hat{i} - 40 t \hat{j}

, and again gives

a = 30 \hat{i} - 40 \hat{j}

, which is the same at every instant. Its magnitude is

∣ a ∣ = 3 0^{2} + 4 0^{2} = 2500 = 50

, so the value at

t = 1

s is

50

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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