Question

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 `hatx` and `hat"y"` directions are defined as 3t and 5t³ respectively, At t = 1s acceleration of the particle will be ______.

Options

• -30 `hat"y"`

• 30 `hat"y"`

• `3hatx+15hat"y"`

• `3hatx+15hat"y"+7hat"z"`

Answer 1

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 `hatx` and `hat"y"` directions are defined as 3t and 5t³ respectively, At t = 1s acceleration of the particle will be 30 `hat"y"`.

Explanation:

We have, `vec"r" = 3"t"hat"i"+5"t"^3hat"j"+7"k"`

So, `("d"^2  vec"r")/("dt"^2)` = 30 t `hat"j" `

At t = 1 ⇒

`("d"^2  vec"r")/("dt"^2)=30hat"j"`

⇒ `vec"a" = 30hat"j"`