Kinematics of a Particle - Speed time graphWatch

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Thread starter 14 years ago
#1
The speed time graph below shows the speed of a particle which accelerates uniformly for 2 second then moves with a constant speed for 6 seconds before retarding to rest.

Find a, the acceleration of the particle b, the retardation of the particle c, the distance it travels during the whole motion.

I am slightly confused what values of V and U to use in order to stick in the formulae of acceleration, thank you v much.
0
14 years ago
#2
Dojo, if you have looked at the book you could see that the difference between the velocity-time graphs and the formulae is that only one of them requires you to use the formulae, now guess which one is that.

On the y-axis you have the velocity, and on the x you have the time. Well U is the INITIAL VELOCITY which means the velocity at time, t=0. Which ever co-ordinate the graph has at the point when t=0 on the y-axis that is your u.

Now for a few BASIC definitions:

Acceleration is defined as the rate of change of velocity, or (dv/dt). So that means that the gradient of a velocity-time graph always gives you the acceleration as long as it is uniform/ linear/ constant for the time period you are measuring it for.

Retardation is like acceleration, the rate of change of velocity but when the particles is SLOWING DOWN. So for (dv/dt) you would expect to get a negative value. Although it is only negative relatively to acceleration. So because you are asked for retardation you give the absolute value of the negative gradient.

Now in practice a velocity-time graph is just a velocity graph. Velocity is defined as (dx/dt), (dx/dt)=v. Now, if you integrate each side with respect to the variable t you get x=vt. This means that the area under a velocity-time graph actually equals to the distance travelled.

Newton.
0
Thread starter 14 years ago
#3
(Original post by Newton)
Dojo, if you have looked at the book you could see that the difference between the velocity-time graphs and the formulae is that only one of them requires you to use the formulae, now guess which one is that.

On the y-axis you have the velocity, and on the x you have the time. Well U is the INITIAL VELOCITY which means the velocity at time, t=0. Which ever co-ordinate the graph has at the point when t=0 on the y-axis that is your u.

Now for a few BASIC definitions:

Acceleration is defined as the rate of change of velocity, or (dv/dt). So that means that the gradient of a velocity-time graph always gives you the acceleration as long as it is uniform/ linear/ constant for the time period you are measuring it for.

Retardation is like acceleration, the rate of change of velocity but when the particles is SLOWING DOWN. So for (dv/dt) you would expect to get a negative value. Although it is only negative relatively to acceleration. So because you are asked for retardation you give the absolute value of the negative gradient.

Now in practice a velocity-time graph is just a velocity graph. Velocity is defined as (dx/dt), (dx/dt)=v. Now, if you integrate each side with respect to the variable t you get x=vt. This means that the area under a velocity-time graph actually equals to the distance travelled.

Newton.
I am sorry, I was just slightly unsure with that question, because I have not done any physics for a while now - time graphs. Yes, I understand what retardation,acceleration,inital velocity , final velocity all is because to even get to that point in the book I had to be able to use them "s,u,v,a,t" g=9.8...v^2=u^2+2as. I will however re-read the time graph tommorrow morning, because I have probably misread something (it is late) as I am trying to get through ASAP so I can revert back to C Mathematics.

0
14 years ago
#4
If you integrate 'g', you get the velocity of 'g', which is the speed of free fall 0
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