Question

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)`

Answer 1

Correct Answer - 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.