Calculating Separation of Cars in a Convoy Moving at Max Speed

[Ujjwal Basumatary]

Homework Statement

In a convoy on a long straight level road, 50 identical cars are at rest in a queue at equal separation $10 m$ from each other. Engine of a car can provide a constant acceleration of $2 m/s^2$. And brakes can provide a maximum deceleration of $4m/s^2$. When an order is given to start the convoy, the first car starts immediately and each subsequent car starts when its distance from a car that is immediately ahead becomes $35m$. Maximum speed limit on this road is $72km/h$. When an order is given to stop the convoy, the driver of the first car applies brakes immediately and the driver of each subsequent car applies brakes with a certain time delay after noticing the brake light of the front car turn red.

1. When all the cars are moving at the maximum speed, what is the separation between two adjacent cars?

There are other parts to the question but I want to attempt them first before requesting a solution.

Homework Equations

$s=ut + \frac{1}{2}at^2$
$v=u+at$
$v^2=u^2+2as$

where $s$, $u$, $v$, $t$ and $a$ denote displacement, initial velocity, final velocity, time and acceleration respectively.

The Attempt at a Solution

Setting up a coordinate system with the origin at the second car, at $t=0$, $S_{1i}=0$ and $S_{2i}=10$ denote respectively the initial positions of the second and the first car at the initial moment. According to the problem statement, the displacement of the first car should be $25m$ from its initial position for the second car to start moving.

Using this fact and the equations of motion I get that at $t=5s$, the first car is at a distance $35m$ from the first. This is the point where the second car starts moving. Taking care of the coordinates and the respective velocity constraints I am getting the separation when both have attained the maximum velocity to be $85m$ whereas the answer given is $110m$. Where have I gone wrong?
Please help.
Thank you.

 

[mfb]

How did you get 85 m? As this is 25 m shorter than the right answer it looks like you forgot that contribution.

How I would calculate it: 5 seconds less at full speed gives 100 m separation, add the 10 m initial separation: 110 m. No need to keep track of the separation as function of time.

 

[PeroK]

Using this fact and the equations of motion I get that at $t=5s$, the first car is at a distance $35m$ from the first. This is the point where the second car starts moving. Taking care of the coordinates and the respective velocity constraints I am getting the separation when both have attained the maximum velocity to be $85m$ whereas the answer given is $110m$. Where have I gone wrong?
Please help.
Thank you.

I just happened to notice that $72km/h = 20m/s$ and that $110m = 10m + (20m/s \times 5s)$. I wonder if that is a coincidence?

 

[Ujjwal Basumatary]

I just happened to notice that $72km/h = 20m/s$ and that $110m = 10m + (20m/s \times 5s)$. I wonder if that is a coincidence?

Got it. Thanks :)

 

[Ujjwal Basumatary]

How did you get 85 m? As this is 25 m shorter than the right answer it looks like you forgot that contribution.

How I would calculate it: 5 seconds less at full speed gives 100 m separation, add the 10 m initial separation: 110 m. No need to keep track of the separation as function of time.

Yeah I missed out that part. Got it anyway. Thanks a lot.