# Quick question about kinematics problem?

• Ishida52134
In summary: Dividing (2) by (3)t1/t2=a2/a2=1/5.Substituting t1/t2=1/5 into (1)(v1)t2=1000v1=500/t2 ....(4)Substituting (4) into (2)500t1/t2=a1t1=0.1t1t1=5000/t2 ...(5)Substituting (4) and (5) into (1)500(5000/t2)/t2=1000t2=√25=5.t1=5000/5=1000.v1=500/5=100.In

## Homework Statement

A commuter train can minimize the time t between two stations by accelerating (a1 = 0.1 m/s^2) for a time t1 then undergoing a negative acceleration (a2 = -0.5 m/s^2) by using his brakes for a time t2. Since the stations are only 1 km apart, the train never reaches its maximum velocity. Find the minimum time of travel t, and the time t1.

## The Attempt at a Solution

basically I listed
a1 = 0.1 m/s^2
a2 = - 0.5 m/s^2
Then I integrated the equations into a piecewise equation for velocity and position.
from 0 < t <= t1 v(t) = t/10
from t1 <= t < t2 v(t) = -t/2 + v(t1)

from 0 < t <= t1 x(t) = t^2 / 20
from t1 <= t < t2 x(t) = -t^2 / 4 + (t1)(t) / 10 + t1^2 / 20

Then I set v(t2) = 0
getting t1 = 5t2.
And I set x(t2) = 1000m using the second equation to solve for t1 and t2.
I eventaully got t1 = 129 s. and t = 154.8 s.

The first piece of the function doesn't equal the second piece at t = t1 so I just want to know what I did wrong. I'm pretty sure I integrated correctly.

thanks.

Ishida52134 said:

## Homework Statement

A commuter train can minimize the time t between two stations by accelerating (a1 = 0.1 m/s^2) for a time t1 then undergoing a negative acceleration (a2 = -0.5 m/s^2) by using his brakes for a time t2. Since the stations are only 1 km apart, the train never reaches its maximum velocity. Find the minimum time of travel t, and the time t1.

## The Attempt at a Solution

basically I listed
a1 = 0.1 m/s^2
a2 = - 0.5 m/s^2
Then I integrated the equations into a piecewise equation for velocity and position.
from 0 < t <= t1 v(t) = t/10
from t1 <= t < t2 v(t) = -t/2 + v(t1)

from 0 < t <= t1 x(t) = t^2 / 20
from t1 <= t < t2 x(t) = -t^2 / 4 + (v(t1))(t) / 10 + t1^2 / 20

Then I set v(t2) = 0
getting t1 = 5t2.
And I set x(t2) = 1000m using the second equation to solve for t1 and t2.
I eventually got t1 = 129 s. and t = 154.8 s.

The first piece of the function doesn't equal the second piece at t = t1 so I just want to know what I did wrong. I'm pretty sure I integrated correctly.

thanks.
Well, it is a bit hard to follow, but t1 should be 5 times t2, which is your result.

The train goes 1000 meters in 154.8 seconds, which is an average velocity of about 6.46 m/s.

So the maximum velocity was twice that. This does give (approximately) the correct accelerations.

Looks good.

For the one you changed, I think it was supposed to be like that. I just substituted v1 into the previous equation making it t1 / 10. Then integrating it I got t1 t /10

Ishida52134 said:
For the one you changed, I think it was supposed to be like that. I just substituted v1 into the previous equation making it t1 / 10. Then integrating it I got t1 t /10
The only (intentional) change I made was putting the v into make it v(t1) and then placing parentheses around that.

You had x(t) = -t^2 / 4 + (t1)(t) / 10 + t1^2 / 20 .

I changed the quantity in red to v(t1), which is what I thought you intended.

yeah that's what I meant lol. v(t1) is the same thing as t1 / 10.
btw, how come for the x(t) piecewise function, the first piece isn't equal to the second at t1? isn't it supposed to be a continuous function.

Ishida52134 said:
yeah that's what I meant lol. v(t1) is the same thing as t1 / 10.
btw, how come for the x(t) piecewise function, the first piece isn't equal to the second at t1? isn't it supposed to be a continuous function.
What do you mean "first piece" ?

0<t<=t1

A commuter train can minimize the time t between two stations by accelerating (a1 = 0.1 m/s^2) for a time t1 then undergoing a negative acceleration (a2 = -0.5 m/s^2) by using his brakes for a time t2. Since the stations are only 1 km apart, the train never reaches its maximum velocity. Find the minimum time of travel t, and the time t1.
---------------------------------

I think there is no minimum time or maximum time of travel.
There is only one maximum velocity thus one unique time.
For other values of time, either it runs short of destination or overshoot.

a1=0.1m/s2
a2=-0.5m/s2
s=1000m
v0=0
v2=0

Unknown, t1,t2,v1

(1/2)v1(t1+t2)=1000 ...(1)
v1=a1t1 ...(2)
v1=-a2t2...(3)

## 1. What is kinematics?

Kinematics is a branch of physics that studies the motion of objects without considering the forces that cause the motion.

## 2. What is a kinematics problem?

A kinematics problem is a question or scenario that involves the study of motion and its related parameters such as position, velocity, and acceleration.

## 3. How do I solve a kinematics problem?

To solve a kinematics problem, you need to identify the known and unknown variables, choose an appropriate kinematics equation, and use algebraic manipulation to find the unknown variable.

## 4. What are some common kinematics equations?

Some common kinematics equations include the equations for calculating displacement (Δx = v0t + 1/2at2), velocity (v = v0 + at), and acceleration (a = Δv/t).

## 5. What are the units of measurement for kinematics?

The units of measurement for kinematics depend on the specific parameter being measured. For example, displacement is measured in meters (m), velocity is measured in meters per second (m/s), and acceleration is measured in meters per second squared (m/s2).