Kinematics

JEE Main level, single correct, medium

39 of 308

Question 39 of 308

The position vector of a particle changes with time according to the relation r⃗(t)=15t2i^+(4−20t2)j^.\vec{r}\left(t\right)=15{t}^{2}\hat{i}+\left(4-20{t}^{2}\right)\hat{j}. What is the magnitude of the acceleration at t=1?t=1?

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Answer: (D) 5050

r⃗=15t2i^+(4−20t2)j^\vec{r}=15{t}^{2}\hat{i}+\left(4-20{t}^{2}\right)\hat{j}
v⃗=30ti^−40tj^\vec{v}=30 t\hat{i}-40t \hat{j}
a⃗=30i^−40j^\vec{a}=30 \hat{i}-40\hat{j}
At t=1t=1 ,
a⃗=30i^−40j^\vec{a}=30 \hat{i}-40\hat{j}
∣a⃗∣=50\left|\vec{a}\right|=50

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