The position vector of a particle related to time {\t} is given by { {\r}}^{\to}= (10 {\t} \stackrel{\^}{ {\i}}+15 {\t}^2 \stackrel{\^}{ {\j}}+7 \stackrel{\^}{ {\k}}) {\m} The direction of net force experienced by the particle is : [15-Apr-2023 shift 1]

Correct Answer is : Positive y-axis

JEE Mains 15-Apr-2023 Shift 1 Solved Paper

Section: Physics

Question : 35 of 90

Marks: +1, -0

The position vector of a particle related to time t is given by

→

=(10t

+15t2

)m
The direction of net force experienced by the particle is :

[15-Apr-2023 shift 1]

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