Question

The position of a body is given as y=t^3+2t^2-6t-4m.Find
1.) the equation of its velocity.
2)it's initial velocity
3)the equation of its acceleration
4.) its initial acceleration
​

Answer 1

Explanation:

differentiation of position gives velocity

differentiation of velocity gives acceleration

y=t^3+2t^2-6t-4 m

therefore, velocity=3t^2+4t-6m/s

at t=0,velocity - initial velocity

v=-6m/s

acceleration=6t+4

at t=0,accleration-initial acceleration

a=4m/s^2