At time t=0 and position s=0 a train pulls out of a station...?

Question

From this point the distance s (in metres) as a function of time t (in seconds) is given by

s = -0.002t^3 + 0.25t^2

i) Determine the velocity, in m.s^-1, after 10 seconds.

ii) Determine the acceleration, in m.s^-2, after 30 seconds.

iii) The velocity begins as 0. The next station is reached when the velocity returns to 0. What time t is this and how far away is the station?

Answer 1

Velocity is the first differential of displacement (distance) and acceleration is the first differential of velocity or the second differential of displacement.
i)V = 3(-0.002)t^2 + 2(0.25)t
V = -0.006t^2 + 0.5t
at t = 10 seconds, V = 4.4ms^-1

ii)A = 2(-0.006)t + 0.5
A = -0.012t + 0.5
at t = 30 seconds, A = 0.14ms^-2

iii)V = 0
0 = -0.006t^2 + 0.5t
0 = t(-0.006t + 0.5)
Therefore t = 0 or 250/3
We are looking for the the latter time so the answer is 250/3 seconds
Using the displacement function:
s = -0.002t^3 + 0.25t^2 and t = 250/3
s = 578.7 meters.
So the next station which is 578.7 meters away takes 250/3 seconds to travel to.

Answer 2

i)
s = -0.002t^3 + 0.25t^2
v = ds/dt = -0.006t^2 + 0.5t
v |t=10 = -0.006(10^2) + 0.5(10) = 4.4 m/s

ii)
a = dv/dt = -0.012t + 0.5
a |t=30 = -0.012(30) + 0.5 = 0.14 m/s2

iii)
v = -0.006t^2 + 0.5t
0 = -0.006t^2 + 0.5t
0.006t^2 - 0.5t = 0
t(0.006t - 0.5) = 0
t = 0, 0.006t - 0.5 = 0
t = 0s, 83 1/3s

s |t=83 1/3 = -0.002(83 1/3)^3 + 0.25(83 1/3)^2 = 578.7 m

The time is 83 1/3 s and the distance is 578.7m away.

Answer 3

s = - 0.002.t³ + 0.25 t²
i)
v(t) = ds/dt = - 0.006.t² + 0.5.t
v(10) = - 0.6 + 5
v(10) = 4.4 m / s (velocity)

ii)
v ` (t) = - 0.012.t + 0.5
v ` (30) = - 0.36 + 0.5
v ` (30) = 0.14 m / s² (acceleration)

iii)
- 0.006 t² + 0.5 t = 0
0.006 t² - 0.5 t = 0
t.(0.006 t - 0.5) = 0
t = 0, t = 0.5 / 0.006
t = 0 ,t = 83.7 secs (to 1 dec. place)
s = - 0.002 (83.7)³ + 0.25 (83.7)²
s = (83.7)² (0.25 - 0.17)
s = 560 m