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Couple of mechanics questions..

Hello just basically some mechanics questions which i cant do right now..

(1) A car is waiting at traffic lights with a van behind it. There is a 1 metre gap between them. When the lights turn green, the car accelerates at 1.5 m/s until it reaches a speed of 15 m/s; it then proceeds at this speed. The van does the same, starting when the gap between the vehicles is 4 metres.
Find a formula for the distance travelled by the car in the first t seconds (0<=t<=10), and hence the time interval between the car starting and the van starting. Find also the distance between the vehicles when they are both going at 15m/s.

(2) A woman stands on the bank of a frozen lake with a dog by her side. She skims a bone across the ice at a speed of 3 m/s. The bone slows down with deceleration 0.4 m/s² , and the dog chases it with acceleration 0.6 m/s². How far out from the bank does the dog catch up with the bone?


Thanks a lot for your time
From John
Reply 1
What specifically can't you do here? These are quite simple problems, so I don't want to just solve them for you as you won't learn anything.
Reply 2
Neo1
Hello just basically some mechanics questions which i cant do right now..

(1) A car is waiting at traffic lights with a van behind it. There is a 1 metre gap between them. When the lights turn green, the car accelerates at 1.5 m/s until it reaches a speed of 15 m/s; it then proceeds at this speed. The van does the same, starting when the gap between the vehicles is 4 metres.
Find a formula for the distance travelled by the car in the first t seconds (0<=t<=10), and hence the time interval between the car starting and the van starting. Find also the distance between the vehicles when they are both going at 15m/s.

(2) A woman stands on the bank of a frozen lake with a dog by her side. She skims a bone across the ice at a speed of 3 m/s. The bone slows down with deceleration 0.4 m/s² , and the dog chases it with acceleration 0.6 m/s². How far out from the bank does the dog catch up with the bone?


Thanks a lot for your time
From John


1a) A formula describing the cars displacement in the first t seconds is simply a constant acceleration equation:





b) If the van starts when the gap between the two vehicles is 4m, it must start when the car has moved 3m, since there is a 1m starting gap. So, the time interval is, using the above equation:






c) Both the car and the van will travel the same distance in accelerating to 15 m/s and take the same time to do so because they both have the same acceleration. Since the van starts 2 s later than the car, once the car reaches 15 m/s it will have travelled for a further 2 s at 15 m/s before the van reaches the same speed. Thus the gap between the cars at this point will be:





2) This is a classic question. By classic I mean that it involves simultaneous equations, which crop up nearly all the time in mechanics! To find the distance at which the dog catches up with the bone, we need to know the time at which this happens. To calculate this time, all we need to realise is that the distances travelled by the dog and the bone must be equal.

So firstly, the distance travelled by the stone as a function of time:





And the same for the distance travelled by the dog:





So, to calculate how long it takes the dog to catch the bone:







Thus, to answer the question, the actual distance travelled out of the bank by both the dog and the bone is, using (2):




Reply 3
Eddie K
1a) A formula describing the cars displacement in the first t seconds is simply a constant acceleration equation:





b) If the van starts when the gap between the two vehicles is 4m, it must start when the car has moved 3m, since there is a 1m starting gap. So, the time interval is, using the above equation:






c) Both the car and the van will travel the same distance in accelerating to 15 m/s and take the same time to do so because they both have the same acceleration. Since the van starts 2 s later than the car, once the car reaches 15 m/s it will have travelled for a further 2 s at 15 m/s before the van reaches the same speed. Thus the gap between the cars at this point will be:





2) This is a classic question. By classic I mean that it involves simultaneous equations, which crop up nearly all the time in mechanics! To find the distance at which the dog catches up with the bone, we need to know the time at which this happens. To calculate this time, all we need to realise is that the distances travelled by the dog and the bone must be equal.

So firstly, the distance travelled by the stone as a function of time:





And the same for the distance travelled by the dog:





So, to calculate how long it takes the dog to catch the bone:







Thus, to answer the question, the actual distance travelled out of the bank by both the dog and the bone is, using (2):






*sigh*
Reply 4
Your problem?
Reply 5
jpowell
What specifically can't you do here? These are quite simple problems, so I don't want to just solve them for you as you won't learn anything.


Oh, no problem :tongue:.
Reply 6
Thanks a lot Eddie K I really appreciate it, I have just been going over it not copying. just a few questions, why do you use for question 1

s = ut + 1/2at² and not any other equation of motion?

Also for question (2) again how did you know you had to use that formula and not any other that has (S) in?

and "So, to calculate how long it takes the dog to catch the bone:" which you got 6 secs. well this just means that the bone had travelled 6 secs too if you substitude to the S(bone) this cant be right, bone has been travelling more time?

Thanks a lot
From John
Reply 7
It's simply a matter of recognising what information you already know. In question 1) we wanted a relationship involving displacement and time, and we have been given an acceleration, so it makes sense to use the equation I did. Using v²=u²+2as wouldn't have got us anywhere because we don't know velocities and the equation doesn't involve time.

For question 2), nowhere in the problem does it say that the dog starts chasing the bone with a delay after the bone is thrown, so we can assume that both the dog and the bone start moving simultaneously. This considered, it is clear that the time for the dog to reach the bone is the same as the time for the bone to be caught by the dog. They are the same thing.

:smile: