# Math Help - Kinematics question

1. ## Kinematics question

A particle moving in a straight line passes through a fixed point O such that it's velocity, v m/s is given by v=6+pt-qt^2, where t is the time in seconds after passing through O, and p and q are constants. when the acceleration is zero, it's velocity is 10 m/s and the particle is 18m from O after 2 seconds.

Find
(i) the values if p and q,
(ii) the interval of t when the velocity is decreasing.

(i) p=6, q=9/4
(ii) t>4/3
Any help please? I differentiate and integrate using the usual methods to find acceleration and displacement respectively. Since the particle is 18m from O, it could mean that s, displacement = both -18m and 18m. Hence, I have two probable equations for the displacement, and 2 e.qs for velocity and acceleration with three unknowns. Then, I'm unable to get anywhere.

A quick answer would be much appreciated, as I need this in time in about 6 more hours. TIA.

A particle moving in a straight line passes through a fixed point, O. It's velocity, v m/s at the time seconds is given by v=2t^2-6t-8. Given that after 3 seconds, the particle is 4m to the left of O.

Find the distance traveled from the beginning until when the acceleration is zero.

2. Originally Posted by smmxwell
Any help please? I differentiate and integrate using the usual methods to find acceleration and displacement respectively. Since the particle is 18m from O, it could mean that s, displacement = both -18m and 18m. Hence, I have two probable equations for the displacement, and 2 e.qs for velocity and acceleration with three unknowns. Then, I'm unable to get anywhere.

A quick answer would be much appreciated, as I need this in time in about 6 more hours. TIA.

For part (1) , maybe you can show your working here so people can check for your mistakes .

Integrate V to get s : $s=6t+\frac{pt^2}{2}-\frac{qt^3}{3}$

Then a=18 when t is 2 , here you get your first equation .

Differentiate V to get a : $a=p-2qt$

Here , when a=0 , $t=\frac{p}{2q}$and at this time , the velocity is 10 .
you get your second equation here . Finally , solve the system .

3. Ops , forgot the other part

For part (1)

(b) When the velocity is decreasing , a<0(slowing down) . Note that it is different from v<0 , this means the object is moving in an opposite direction .

So when you have the values of p and q , plug them into a and find the time intervals .

4. Yeah. I tried your method beforehand, but I still can't get the answer. I get a different answer than the book. It could be since it's quite late here, that I might have made some careless mistakes.

For my working,
I get:
a=p-2qt

From s, 9= 3p-4q or -45=3p-4q (as displacement could be either positive or negative, which isn't stated in the question.)

When v=10, 4=pt-qt^2

And then I just start substituting the equivalents into each eq, but to no avail.

5. Okay. I managed to get the answer.