JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

1 Easy

1 of 1

Application +4 | -1

A particle moves along a curve so that its coordinates at time

t

are

x=t, y=\frac{1}{2} t^2, z=\frac{1}{3} t^3

. The acceleration at

t=1

View answer and explanation

Correct option: A

{j}+2 {k}

Option B (

{j}+2 {k}

) is correct because the acceleration vector at t=1 is found by taking the second derivative of the position vector and evaluating it at t=1.

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O A

O B

30^{\circ}

45^{\circ}

O B

2x + 4y - 5z = 10