How Fast Does the Train Accelerate to Outpace the Car?

[theultimate6]

Homework Statement

A 108 m long train starts from rest (at t = 0) and accelerates uniformly. At the same time (at t = 0), a car moving with constant speed in the same direction reaches the back end of the train. At t = 12 s the car reaches the front of the train. However, the train continues to speed up and pulls ahead of the car. At t = 32 s, the car is left behind the train. Determine,
a. the car’s speed
b. the trains acceleration

The Attempt at a Solution

When the car reached the front of the train it moved a distance of [tex]x+108[/tex] where x is the distance covered by the train

from [tex]x=Vot+0.5at^2[/tex]

[tex]x=0.5at^2 =0.5*a*12^2 =72a[/tex]

the distance covered by the car in the first 12 seconds is [tex]72a+108[/tex]

that means its speed is
[tex]v=x/t = (72a+108)/12 =6a+9[/tex]

by 32 seconds the car has moved a distance of

[tex]x=vt[/tex]
[tex]=(6a+9) *(32-12) =20(6a+9)=120a+180[/tex]

the train covered this distance + 108 meters:
distance covered by the train : [tex]120a+180+108 =120a+288[/tex]

the distance covered by the train is also

[tex]V+0.5a*20^2+108[/tex]

Where [tex]V= 72a[/tex]

Therefore : [tex]72a+0.5a*20^2+108=120a+288[/tex]

Solving for [tex]a[/tex] we get

[tex]a=0.5263 m/s^2[/tex]

therefore [tex] V_c =12.1578 m/s [/tex]


Not really sure about this solution can anyone correct me

 

[SammyS]

Homework Statement

A 108 m long train starts from rest (at t = 0) and accelerates uniformly. At the same time (at t = 0), a car moving with constant speed in the same direction reaches the back end of the train. At t = 12 s the car reaches the front of the train. However, the train continues to speed up and pulls ahead of the car. At t = 32 s, the car is left behind the train. Determine,
a. the car’s speed
b. the train's acceleration

The Attempt at a Solution

When the car reached the front of the train it moved a distance of [tex]x+108[/tex] where x is the distance covered by the train

from [itex]\ \ x=V_0 t+0.5at^2[/itex]
[tex]x=0.5at^2 =0.5*a*12^2 =72a[/tex]the distance covered by the car in the first 12 seconds is [tex]72a+108\ .[/tex]
That means its speed is [tex]v=x/t = (72a+108)/12 =6a+9\ .[/tex]

What you have below occurs during the time from the 12 second mark to the 32 second mark.

by 32 seconds the car has moved a distance of [tex]x=vt[/tex][tex]=(6a+9) *(32-12) =20(6a+9)=120a+180[/tex]the train covered this distance + 108 meters:
distance covered by the train : [tex]120a+180+108 =120a+288[/tex]the distance covered by the train is also
[tex]V+0.5a*20^2+108[/tex]

You should not have added the 108 in the above equation.

What you have for the following velocity is actually the distance the train traveled during the first 12 seconds.

The train's velocity at the 12 second mark is the same as the car's velocity (Do you know why?) and is given by [itex]\ \ V=6a+9\ .[/itex]

Where [tex]V= 72a[/tex]
Therefore : [tex]72a+0.5a*20^2+108=120a+288[/tex]
Solving for [tex]a[/tex] we get [tex]a=0.5263 m/s^2[/tex]
therefore [tex] V_c =12.1578 m/s [/tex]
Not really sure about this solution can anyone correct me

Instead of considering the situation from the 12 second mark to the 32 second mark, you could have ised the fact that from the 0 second mark to the 32 second mark, both the car and train travel the same distance.