Calculus: Particle velocity - 10 POINTS!

A particle moves in a line with position at time t given by 30t − 4t^2
(a) Find the average velocity for the interval of length 0.5 sec ending at t = 3
(b) Find the velocity at the instant t = 3 using limits
(c) Find the velocity at the instant t = 0 using limits (hint: the answer is not 0!)
(d) find the velocity the instant t = a using limits
(e) Find the time at which the velocity is 0.

(a)
Average Velocity=( Total Distance Traveled )/( Total Time Taken )
=[integral (30t − 4t^2) t=0.5 to t=3]/(3-0.5)
=38.1667

(b)
Velocity
=dx/dt
=Lt h->0 [{30(3+h)-4(3+h)^2}-{30*3-4*3^2}]/(3+h-3…
=30-8*3
=6m/s

(c)
Velocity
=dx/dt
=Lt h->0 [{30(0+h)-4(0+h)^2}-{30*0-4*0^2}]/(h+0-0…
=30-8*0
=30m/s

(d)
Velocity
=dx/dt
=Lt h->0 [{30(a+h)-4(a+h)^2}-{30*a-4*a^2}]/(h+a-a…
=30-8*a

(e)
Velocity=0
=> dx/dt=0
=> Lt h->0 [{30(t+h)-4(t+h)^2}-{30*t-4*t^2}]/(h+t-t…
=> 30-8t=0
=> t=30/8
=> t=3.75 seconds

(a) 8 units/sec
(b) 6 units/sec
(c) 30 units/sec
(d) (30 - 8a) units/sec
(e) t = 3.75

hope this helps :)

For part a, i think you mean a time interval of 0.5 sec? Which means t=2.5 to t=3
So {30(3)-4(3)^2 - [30(2.5)-4(2.5)^2] } divided by 0.5 = 8 units/sec
As for the rest you can check out Nehal J's answer.