MHB Mechanics- General motion in a straight line.

AI Thread Summary
The discussion focuses on calculating the distance a goods train travels from point A to point B, considering its acceleration, constant velocity, and deceleration phases. The acceleration is defined by the equation a=0.1t^2(6-t) for 0<t<6, leading to a calculated distance of 1690 meters, as confirmed by integration and kinematics equations. Participants discuss their methods, including integration and kinematic formulas, to arrive at the distance traveled during each phase of motion. Clarifications are sought regarding the calculation of distance during deceleration, with specific emphasis on the average velocity approach. The final consensus is that the total distance from A to B is 1690 meters to three significant figures.
Shah 72
MHB
Messages
274
Reaction score
0
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.
 
Mathematics news on Phys.org
$\displaystyle d = \int_0^6 v(t) \, dt + v(6) \cdot 150 + 4.5[v(156)+v(165)]$

$d=1690$m to three significant figures
 
skeeter said:
$\displaystyle d = \int_0^6 v(t) \, dt + v(6) \cdot 150 + 4.5[v(156)+v(165)]$

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

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

$d=1690$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?
 
skeeter said:
$\displaystyle d = \int_0^6 v(t) \, dt + v(6) \cdot 150 + 4.5[v(156)+v(165)]$

$d=1690$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
 
skeeter said:
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!
 

Similar threads

Replies
4
Views
2K
Replies
2
Views
1K
Replies
4
Views
966
Replies
7
Views
895
Replies
2
Views
964
Replies
5
Views
1K
Replies
8
Views
1K
Back
Top