# Mechanics- General motion in a straight line.

• MHB
• Shah 72

#### Shah 72

MHB
A goods train starts from rest at point A and moves along a straight track. The train moves with acceleration a m/s^2 at time t s, given by a=0.1t^2(6-t) for 0<t<6. It then moves at constant velocity for 6<t<156 before decelerating uniformly to stop at point B at t=165. Calculate the distance from A to B.
I have no clue how to solve this.
I used the usual method without integration and got s=1700m. The textbook ans says 1690.

$\displaystyle d = \int_0^6 v(t) \, dt + v(6) \cdot 150 + 4.5[v(156)+v(165)]$

$d=1690$m to three significant figures

$\displaystyle d = \int_0^6 v(t) \, dt + v(6) \cdot 150 + 4.5[v(156)+v(165)]$

$d=1690$m to three significant figures
Thank you!

$\displaystyle d = \int_0^6 v(t) \, dt + v(6) \cdot 150 + 4.5[v(156)+v(165)]$

$d=1690$m to three significant figures

Can I pls ask you about the last part, how you got 4.5 ( v156+v165) do I need to integrate?

$\displaystyle d = \int_0^6 v(t) \, dt + v(6) \cdot 150 + 4.5[v(156)+v(165)]$

$d=1690$m to three significant figures
I don't understand the last part.
I get s1 for the interval between 0 and 6= 25.92m
Constant velocity, v=s/t, I get s2 for interval between 6 and 156= 1620m.

for the period of uniform deceleration, $\Delta t = 9$ sec.

recall the kinematics equation for uniform acceleration …

$\Delta x = v_{avg} \cdot \Delta t = \dfrac{1}{2}(v_0 + v_f) \cdot \Delta t = 4.5(10.8 + 0) = 48.6$ m

you could also use …

$\Delta x = v_0 \Delta t + \dfrac{1}{2}at^2 = 10.8 \cdot 9 - 0.6 \cdot 9^2 = 48.6$ m

for the period of uniform deceleration, $\Delta t = 9$ sec.

recall the kinematics equation for uniform acceleration …

$\Delta x = v_{avg} \cdot \Delta t = \dfrac{1}{2}(v_0 + v_f) \cdot \Delta t = 4.5(10.8 + 0) = 48.6$ m

you could also use …

$\Delta x = v_0 \Delta t + \dfrac{1}{2}at^2 = 10.8 \cdot 9 - 0.6 \cdot 9^2 = 48.6$ m
Thanks a lot!