# Math Help - Variable acceleration

1. ## Variable acceleration

Hi I just wondered if what i have done is correct. before i carry on doing these questions.

1. A vehicle, moving along a straight horizontal road , has velocity v m/s at time t seconds modelled by

$v = \frac{1}{60}t(60-t)$ $0<=t<=60$

a) write down the times at which the vehicle is stationary

b) calculate the distance travelled by the vehicle between t=0 and t=60

c) i) Find, in terms of t, an expression for the acceleration of the vehicle at time t seconds

ii) Find the greatest speed of the vehicle

a) vehicle startionary at t=0 and t=60

b) $x=\int_0^{60} \frac{dx}{dt}$

$\frac{1}{60}\int_0^{60} 60t-t^2dt$

= $\frac{1}{60}(30t^2 - \frac{1}{3}t^3)_0^{60}$

= $\frac{1}{60}(30(60^2)-\frac{1}{3}(60^3))$

=600m

c) $v = t - \frac{t^2}{60}$

$\frac{dv}{dt} = 1 - \frac{2t}{60}$

greatest speed when a = 0

$1 - \frac{2t}{60} = 0$

$60 - 2t = 0$
2t = 60
t = 30s

when t=30

$v = \frac{1}{60}(30)(60-30)$

$= \frac{900}{60} = 15m.s$

2. a.) True when v(t) is 0, as you've shown at 0s and 60s

b.) Integrate for s(t), initial distance is 0 so C = 0, answer is 600m

c.) i.) v'(t) = a(t) ii.) occurs when a(t) = 0 and we go from positive to negative. True at t=30

Looks good to me.