At time t=0 a particle starts travelling from a height 7z^ cm in a plane keeping z coordinate constant. At any instant of time, it's position along…

Question

At time t=0 a particle starts travelling from a height 7z^ cm in a plane keeping z coordinate constant. At any instant of time, it's position along the x and y directions are defined as 3t and 5t^3 respectively. At t=1 s acceleration of the particle will be

MARKS Content Team · Accepted Answer

Correct answer: (B) 30y. The position vector of the particle can be written as, r→=3ti^+5t^3j^+7k^ The velocity of the particle will be, v→=(dr→)/dt=3i^+15t^2j^ Now the acceleration of the particle will be, a→=(dv→)/dt=(d^2r→)/(dt^2)=30tj^ At t=1 s⇒(d^2r)/(dt^2)=30j^

At time $t = 0$ a particle starts travelling from a height $7 \hat{z} cm$ in a plane keeping $z$ coordinate constant. At any instant of time, it's position along the $x$ and $y$ directions are defined as $3 t$ and $5 t^{3}$ respectively. At $t = 1 s$ acceleration of the particle will be

Solution

Relevant Questions

At time t=0 a particle starts travelling from a height 7z^ cm in a plane keeping z coordinate constant. At any instant of time, it's position along the x and y directions are defined as 3t and 5t3 respectively. At t=1 s acceleration of the particle will be

Solution

Relevant Questions

At time $t = 0$ a particle starts travelling from a height $7 \hat{z} cm$ in a plane keeping $z$ coordinate constant. At any instant of time, it's position along the $x$ and $y$ directions are defined as $3 t$ and $5 t^{3}$ respectively. At $t = 1 s$ acceleration of the particle will be