Question

The position x of a partical varies time t as x=at square -bt cube where a and b are constant. What is the velocity of the particle at t=1s

Answer 1

Answer:

2a-3b

Explanation:

Differentiating w. r. t. x, we get

dx/dt= 2at-3bt^2

•°•v= 2at-3bt^2

At t=1 ,

v= 2a-3b