calc answer in 20 minutes?

Question

The acceleration of a particle moving along the coordinate line is 5+4t m/s^2. At t=0 the velocity is 5 m/s. Find how far the particle moves from time t=0 to t=2.

thanks in advance!!

Answer 1

42Metres
at t=0 v=5m/s
at t=2
v=u-at
=5-(5+4*2)
=8m/s

d=vt+0.5at^2
=8*2+.5*(5+4*2)*2^2
=42m

Answer 2

acceleration = 5 + 4t

so velocity = 5t + 2t^2 + c1

at t = 0, v = 5
so c1 = 5
so velocity = 5t + 2t^2 + 5

so displacement = (5/2)t^2 + (2/3)t^3 + 5t + c2

so total distance = (displacement at t=2) - (displacement at t=0) = (5/2)*2^2 + (2/3)*2^3 + 5*2 + c2 - c2 = 10 + 16/3 + 10 = 76/3 = 25.33 m

So the particles moves 25.33 meters from t=0 to t=2

Answer 3

dv/dt=5+4t
v = 5t + 2t^2 + c
to find c, substitute t=0 and v=5 ms^-1
then,
5= 5(0) +2(0)+c => c=5
then,
ds/dt = 5 + 5t + 2t^2
s = [5t + (5/2)t^2 + (2/3)t^3]0 to 2
= {5(2) + 10 + 16/3} - {0}
= 76/3 m

Answer 4

5 m/s + 4 * 2s * m/s^2 = 13 m/s

come on, was this hard???