Kinematics Assistance (URGENCY 2.0)
I will give a few questions that I need help with.
A car is initially travelling with a constant velocity of 15 ms-1 for T s. It then decelerates at a constant rate of T/2 s , reaching a velocity of 10 ms-1. It then immediately accelerates at a constant rate of 3T/2 s, reaching a velocity of 20 ms-1.
a) sketch a velocity time graph to illustrate motion (DONE)
*b) given that the car travels a total distance of 112,5 m over the journey described, find the value of T.*
I'm not sure if my answer is correct. I initially got 9.72, but I'm not certain. I think something went wrong; tried to fix the problem but got flustered.
2. A racing car starts from rest at the point A, and moves with constant acceleration of 11ms-2 for 8 seconds. The velocity it has reached after 8 s is then maintained for T s. The racing car then decelerates from this velocity to 40 ms-1, in a further 2 s , reaching point B.
A) sketch a velocity-time graph (done)
*b) Given that the distance between A and B is 1404 m , find the value of T.*
3. A cyclist is descending down a mountain with constant acceleration. She passes through three checkpoinnts, P, Q and R, with velocity 6 ms-1, x ms-1 and 20 ms-1 respectively. Time taken to trvel from P to R = 35 s.
A) Find acceleration
I got the acceleration to equal 0.4 ms-2, I'm sure but does anyone mind to confirm this, if wrong, please explain.
Given that t1/t2 = 4/3 , where t1 is time taken to travel from P to Q and t2 is the time taken to travel from Q-R,
*b) find the value of x*
I didn't understand this, I need a lot of help on this one
c) Find the distance between P and R. I can do this one, but it's one of these continuous questions and I don't want an ecf.
Please help me, with the questions requested, Thank you so much!!!!!! I really appreciate it.
A car is initially travelling with a constant velocity of 15 ms-1 for T s. It then decelerates at a constant rate of T/2 s , reaching a velocity of 10 ms-1. It then immediately accelerates at a constant rate of 3T/2 s, reaching a velocity of 20 ms-1.
a) sketch a velocity time graph to illustrate motion (DONE)
*b) given that the car travels a total distance of 112,5 m over the journey described, find the value of T.*
I'm not sure if my answer is correct. I initially got 9.72, but I'm not certain. I think something went wrong; tried to fix the problem but got flustered.
2. A racing car starts from rest at the point A, and moves with constant acceleration of 11ms-2 for 8 seconds. The velocity it has reached after 8 s is then maintained for T s. The racing car then decelerates from this velocity to 40 ms-1, in a further 2 s , reaching point B.
A) sketch a velocity-time graph (done)
*b) Given that the distance between A and B is 1404 m , find the value of T.*
3. A cyclist is descending down a mountain with constant acceleration. She passes through three checkpoinnts, P, Q and R, with velocity 6 ms-1, x ms-1 and 20 ms-1 respectively. Time taken to trvel from P to R = 35 s.
A) Find acceleration
I got the acceleration to equal 0.4 ms-2, I'm sure but does anyone mind to confirm this, if wrong, please explain.
Given that t1/t2 = 4/3 , where t1 is time taken to travel from P to Q and t2 is the time taken to travel from Q-R,
*b) find the value of x*
I didn't understand this, I need a lot of help on this one
c) Find the distance between P and R. I can do this one, but it's one of these continuous questions and I don't want an ecf.
Please help me, with the questions requested, Thank you so much!!!!!! I really appreciate it.
Scroll to see replies
For the first question, part b, you can produce the equation 15T + ((10+15)/2) * T/2 + ((20+10)/2) * 3t/2 = 112.5 assuming you mean accelerates at a constant rate for the given times, I get the (10+15)/2 and (20+10)/2 as this gives the average velocity for the given time since acceleration is constant and then multiply each of the velocities by the time that the car is travelling for which gives distance which is given, then you solve this equation which gives T = 2.5714s
Original post by MattSull
For the first question, part b, you can produce the equation 15T + ((10+15)/2) * T/2 + ((20+10)/2) * 3t/2 = 112.5 assuming you mean accelerates at a constant rate for the given times, I get the (10+15)/2 and (20+10)/2 as this gives the average velocity for the given time since acceleration is constant and then multiply each of the velocities by the time that the car is travelling for which gives distance which is given, then you solve this equation which gives T = 2.5714s
The asterisks represent multiplication, right?
Just checking.
For question 2b, we can find the distance travelled during the first section, when the racing car is accelerating, we know it is accelerating for 8 seconds at a rate of 11ms-2 from a velocity of 0ms-1 so that means that the final velocity here is v=u+at=0+11*8=88ms-1 so the distance travelled is s=(u+v)*t/2= (0+88)*8/2=352m, then during the second part, the velocity is 88ms-1 for Ts so that gives d=v*t=88T and then for the final part, we know that the initial velocity is 88ms-1 and the final is 40ms-1 which is achieved in 2s so use s=(u+v)*t/2 again giving s=0.5(88+40)*2=88+40=124, now combine these so 352+88T+124=1404 so 88T=928 so T=928/88=10.545454...
And finallyy for question 3, part A can be found using the initial velocity (u) of 6ms-1 and the final velocity (v) of 20 and time (t) = 35, rearrange the equation v=u+at to give a=(v-u)/t=(20-6)/35=0.4ms-2 as the change in velocity is 20-6 and acceleration is change in velocity over time so correct there, for part b, you know that t1+t2=35 as that is the total time from P to R and you have the equation t1/t2=4/3, rearrange the second to get it in the form a*t1+b*t2=c as such: t1=4*t2/3 so 3*t1=4*t2 so 3*t1+4*t2=0, now the two equations are in the same form they are easy to solve simultaneously, giving t1 = 20 and t2 = 15 so using v=u+at where u=6, a=.4 and, t=t1=20, v=6+.4*20=14ms-1=x and I will leave you to do part c, just us s=(v+u)*t/2
Any questions about that feel free to ask.
Any questions about that feel free to ask.
Original post by KatheO11
The asterisks represent multiplication, right?
Just checking.
Just checking.
Yes sorry they do represent multiplication
Original post by MattSull
so using v=u+at where u=6, a=.4 and, t=t1=20, v=6+.4*20=14ms-1=x and I will leave you to do part c, just us s=(v+u)*t/2
Any questions about that feel free to ask.
so using v=u+at where u=6, a=.4 and, t=t1=20, v=6+.4*20=14ms-1=x and I will leave you to do part c, just us s=(v+u)*t/2
Any questions about that feel free to ask.
The thing about the v = u+at, I got 86 ms -1 = 14 ms-1 , is that what you meant or did I do something wrong? I didn't comprehend that bit.
Other than that, You are such an Angel! Thank you so much!I greatly appreciate the assistance you have given to me.
Original post by KatheO11
The thing about the v = u+at, I got 86 ms -1 = 14 ms-1 , is that what you meant or did I do something wrong? I didn't comprehend that bit.
Other than that, You are such an Angel! Thank you so much!I greatly appreciate the assistance you have given to me.
Other than that, You are such an Angel! Thank you so much!I greatly appreciate the assistance you have given to me.
What numbers did you use in the equation?
Original post by MattSull
What numbers did you use in the equation?
ah I see, sorry my bad habit, I put .4 instead of 0.4, 0.4 is the acceleration we found in the first part of the question, using that it does make 14
Original post by MattSull
What numbers did you use in the equation?
u = 6
a = 4
t = 20.
Oh, ummm, I need help with another question.
4) A particle P passes through Point A with velocity 2.8 ms-1 and constant acceleration 0.12 ms-2. Three seconds laster a second particle Q passes through A with velocity 2.4 ms-1 and constant acceleration 0.2 ms-2.
a) write down the expressions for the displacements of P and Q from A, in terms of t, where t s is the time after P passed through A.
b) show that, when the particles meet, 2t^2 - 50 t -315 = 0
c) find the distance when 2 particles meet.
Okay for question 4, we start out with writing out all of the information that we have, (I am using 1 to represent values for particle P and 2 to represent values for particle Q) u1 = 2.8, a1 = 0.12 and t1 = T as well as u2 = 2.4, a2 = 0.2 and t2 = T-3 now we want to create an expression for each of the displacements (s) we get from A at a given time T, the SUVAT equation which includes each of these values is s=ut+1/2at, so for particle P we get s1=2.8T+.06T^2 and for particle Q we get s2=2.4(T-3)+.1(T-3)^2 where ^2 represents to the power of 2 these are your two equations for displacement, the second equation still needs simplifying from here. For part b of this question we are saying that the displacement of both particles are equal (i.e. they meet) so s1=s2, I will leave you to do this, you just need to make the two equations equal each other and this should give the equation from the question, for the final question you just need to solve that equation and it should give the values of T where the particles meet and then use these values to find the different displacements which they meet at.
(edited 6 years ago)
Original post by MattSull
Okay for question 4, we start out with writing out all of the information that we have, (I am using 1 to represent values for particle P and 2 to represent values for particle Q) u1 = 2.8, a1 = 0.12 and t1 = T as well as u2 = 2.4, a2 = 0.2 and t2 = T-3 now we want to create an expression for each of the displacements (s) we get from A at a given time T, the SUVAT equation which includes each of these values is s=ut+1/2at, so for particle P we get s1=2.8T+.06T^2 and for particle Q we get s2=2.4(T-3)+.1(T-3)^2 where ^2 represents to the power of 2 these are your two equations for displacement, the second equation still needs simplifying from here. For part b of this question we are saying that the displacement of both particles are equal (i.e. they meet) so s1=s2, I will leave you to do this, you just need to make the two equations equal each other and this should give the equation from the question, for the final question you just need to solve that equation and it should give the values of T where the particles meet and then use these values to find the different displacements which they meet at.
Again, thank you so much!!!!!
Original post by KatheO11
Again, thank you so much!!!!!
No problem, glad I could help
Original post by MattSull
Okay for question 4, we start out with writing out all of the information that we have, (I am using 1 to represent values for particle P and 2 to represent values for particle Q) u1 = 2.8, a1 = 0.12 and t1 = T as well as u2 = 2.4, a2 = 0.2 and t2 = T-3 now we want to create an expression for each of the displacements (s) we get from A at a given time T, the SUVAT equation which includes each of these values is s=ut+1/2at, so for particle P we get s1=2.8T+.06T^2 and for particle Q we get s2=2.4(T-3)+.1(T-3)^2 where ^2 represents to the power of 2 these are your two equations for displacement, the second equation still needs simplifying from here. For part b of this question we are saying that the displacement of both particles are equal (i.e. they meet) so s1=s2, I will leave you to do this, you just need to make the two equations equal each other and this should give the equation from the question, for the final question you just need to solve that equation and it should give the values of T where the particles meet and then use these values to find the different displacements which they meet at.
would t2 not be t 3?
Original post by MattSull
And finallyy for question 3, part A can be found using the initial velocity (u) of 6ms-1 and the final velocity (v) of 20 and time (t) = 35, rearrange the equation v=u at to give a=(v-u)/t=(20-6)/35=0.4ms-2 as the change in velocity is 20-6 and acceleration is change in velocity over time so correct there, for part b, you know that t1 t2=35 as that is the total time from P to R and you have the equation t1/t2=4/3, rearrange the second to get it in the form a*t1 b*t2=c as such: t1=4*t2/3 so 3*t1=4*t2 so 3*t1 4*t2=0, now the two equations are in the same form they are easy to solve simultaneously, giving t1 = 20 and t2 = 15 so using v=u at where u=6, a=.4 and, t=t1=20, v=6 .4*20=14ms-1=x and I will leave you to do part c, just us s=(v u)*t/2Any questions about that feel free to ask.
Original post by help please765
i got confused where the simultaneous equations got involved. How did you re arrange the equation and which two equations were use din the sim equation to get 20. PLEASE HELP
I assume you're refering to
Given in the question, we have:
Multiply up, giving.
and rearrange to
........................(1)
We also know the total time taken is 35s, so:
.........................(2)
Those are the two equations, and solve via whatever method you prefer.
Thank you... really appreacited
really appreaciated .
for question 1b what is the time that i multiply by each velocity...... IMMA START CRYING ...
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