Question

. A car moves in a straight line such that for a short time its velocity is defined by
v = (5t2 + 3t) ft/s, where ‘t’ is in seconds. Determine its position and acceleration when t = 4s. When t = 0, s = 0.

Answer 1

Answer:

a=dv/dt

a= 10t+3

put t = 4

a= 43m/s^2

dx/dt= v

dx = vdt

dx = (5t^2 + 3t)dt

x= 5/3t^3 + 3/2t^2

Now Put Limit t=0 x=0 & t=4 x=X

x-0 = 320/3 + 24

x= 392/3 m