Question

the position of an object moving along a straight line is given by equation x= 18t+5t^2. find the velocity of the object when t=0s and t=3s. also find the average velocity for time interval t=2s and t=4s​

Answer 1

Answer:

x=5t

2

+18t

(1)v=

dt

dx

=10t+18

att=2s,v=10×2+18=38m/s

(2)Average,velocity=

time

displacement

att=2,x=5(2)

2

+18×2=56m

t=3,x=5(3)

2

+18×3=99m

Averagevelocity=

3−2

99−56

=43 m/s