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In the question number 35, the accelerat...

In the question number 35, the acceleration of the particle at `t = 1` is

A

`2hatjm s^(-2)`

B

`-2 hatj ms^(-2)`

C

`4 hatj ms^(-2)`

D

`-4hatj ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
C
From the previous question, `vecv = 3hati + 4t hatj ms^(-1)`
`therefore ` Acceleration, `veca = (dvecv)/(dt) = 4hatj ms^(-2)`
Acceleration of the particle remains constant at all times.
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