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

1.	In the question number 35, the acceleration of the particle at t = 1 is
Answer» `2hatjm s^(-2)` `-2 hatj ms^(-2)` `4 hatj ms^(-2)` `-4hatj ms^(-2)` Solution :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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