Question

12. A train starts from a station with a constant acceleration of at = 0.40 m/s2.

A passenger arrives at the track time t = 6.0s after the end of the train left the very same point.

What is the slowest constant speed at which she can run and catch the train? Sketch curves for the motion of the pasenger and the train as functions of time.

Answer 1

To catch the train accelerating at 0.40 m/s², the passenger must run with a minimum constant speed of 1.2 m/s. This is calculated based on the distance the train travels in the first 6 seconds, which the passenger must be able to cover at the same time interval.

Step-by-step explanation:

Calculating the Minimum Speed to Catch the Train

To find the slowest constant speed at which the passenger can run and catch the train that started accelerating from the station with an acceleration of 0.40 m/s2, we need to calculate the distance the train will cover during those 6 seconds and ensure that the passenger's speed allows them to cover at least the same distance in the same time.

The distance the train covers in 6 seconds can be calculated using the kinematic equation for uniformly accelerated motion:

• d = (1/2) * a * t2

Where:

• a is acceleration
• t is time

Substituting the given values we get:

d = (1/2) * 0.40 m/s2 * (6 s)2 = 7.2 m

The passenger must cover this 7.2 meters in 6 seconds to catch the train. The minimum constant speed v necessary is therefore:

v = d / t = 7.2 m / 6 s = 1.2 m/s

The passenger needs to run with a minimum constant speed of 1.2 m/s to catch up with the train.

For the graphical representation, both the train's and passenger's distance versus time graphs should start from the same point, but at t = 6s, the train's graph will show a parabolic curve due to acceleration, and the passenger's graph will show a straight line if running at a constant speed.

Answer 2

If the passenger instantly gets to this minimal speed to catch the train, his speed would be 4.8m/s

If we consider our initial time when the passenger start running, we can describe the motion equations:

`X_(pas)=v_0t`

`X_(train)=(1)/(2) 0.4(m)/(s^2)(t+6s)^2`

The minimum passenger speed will be the one that makes these 2 curves intersect in a single point. In other words, the motion curve of the passenger must be the tangent curve of the train motion curve that crosses for the point (0,0).

The tangent curve of a function f(t) is defined:

`T(t)=f'(a)(t-a)+f(a)` where t=a is the time at which both curves cross.

In this case:

`T(0)=X_(pas)(0)=0=f'(a)(-a)+f(a)`

`0=f'(a) \cdot(-a)+f(a)` and
`f(t)=X_(train)(t)`

`X_(train)(a)=(1)/(2) 0.4(m)/(s^2)(a+6s)^2\\X_(train)'(a)=(1)/(2) 0.4(m)/(s^2)(2a+12s)\\-(1)/(2) 0.4(m)/(s^2)(2a+12s)a+(1)/(2) 0.4(m)/(s^2)(a+6s)^2 \leftrightarrow a=6s`.

Therefore:

`X_(pas)=v_0t=f'(6s)(t-6s)+f(6s)=4.8(m)/(s)t`