Question

Given the velocity function v(t) = x^2+2x+1

a. Determine the position function
b. Determine the acceleration function
c. What is the position, velocity and acceleration of the particle at t=5s?

Answer 1

Answer:

The displacement is given as x=5t

2

+4t+3

⇒x−3=5t

2

+4t

assuming x−3=s,

s=4t+

2

1

10t

2

comparing this with s=ut+

2

1

at

2

we get u=4m/s and a=10m/s

2

so velocity at t=2sec is v=u+at=4+10×2=24m/s

Answer 2

Answer:

instead of X , you can put t as variable..

hope it helps you.... stay blessed:)