Mechanics- General motion in a straight line.

In summary, the conversation discusses the calculation of the distance from point A to point B for a goods train that starts from rest and undergoes acceleration, constant velocity, and uniform deceleration. The distance is calculated using integration and kinematics equations, resulting in a distance of 1690 meters to three significant figures.
  • #1
Shah 72
MHB
274
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.
 
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  • #2
$\displaystyle d = \int_0^6 v(t) \, dt + v(6) \cdot 150 + 4.5[v(156)+v(165)]$

$d=1690$m to three significant figures
 
  • #3
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!
 
  • #4

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?
 
  • #5
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.
 
  • #6
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
 
  • #7
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!
 

Related to Mechanics- General motion in a straight line.

1. What is the difference between speed and velocity?

Speed is a measure of how fast an object is moving, while velocity is a measure of both the speed and direction of an object's motion.

2. What is the equation for calculating average speed?

The equation for average speed is: speed = distance / time.

3. How is acceleration defined in mechanics?

Acceleration is the rate at which an object's velocity changes over time. It is defined as the change in velocity divided by the change in time.

4. What is the difference between uniform and non-uniform motion?

Uniform motion is when an object moves at a constant speed, while non-uniform motion is when an object's speed changes over time.

5. How does Newton's first law of motion relate to mechanics?

Newton's first law of motion states that an object will remain at rest or in motion with a constant velocity unless acted upon by an external force. This law is essential in understanding the behavior of objects in motion in mechanics.

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