# Thread: Need help with a kinematics question

1. ## Need help with a kinematics question

Hi, I have a kinematics question which i can't solve and i need help. Here it is.

A particle moves in a straight line passing a fixed point O with a velocity of 3m/s. It moves in such a manner that t seconds after passing O, its velocity is given by v= atē+b. If the particle is again at O after 3 seconds, find its speed at that instant. Find the total distance traveled between t=0 and t=3.

This is what i have tried so far :
v= atē+b

subst. t=0, v=3.
3 = a(0)ē + b
b=3

∴ subst. b=3, t=3 into v= atē+b
v = a(3)ē + 3
v = 9a + 3
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I couldn't continue after arriving here, can't get the unknown,a. I only need help here. I understand how to get the answer for the distance part. Thanks

2. As far as I see, the problem involves 2 unknowns with 1 equation!!

3. Ya! I've been thinking for a long time to get the other unknown

4. I don't understand the quoted part of your question:

Originally Posted by Unagi9
...
A particle moves in a straight line passing a fixed point O

If the particle is again at O
...
In my opinion the particle can pass a certain place several times if it moves on a closed curve (circle, ellipse, ...). But then it doesn't move in a straight line.

Or: If the particle moves in a straight line it passes a certain place only once.

5. Originally Posted by Unagi9
A particle moves in a straight line passing a fixed point O with a velocity of 3m/s. It moves in such a manner that t seconds after passing O, its velocity is given by v= atē+b. If the particle is again at O after 3 seconds, find its speed at that instant. Find the total distance traveled between t=0 and t=3.
the particle changes direction as it moves in a straight line ...

$v(0) = 3 \, m/s$

$v(t) = at^2 + b$

$b = 3$

since the particle returns to point O at $t = 3$ , then its displacement from $t = 0$ to $t = 3$ is zero ...

$\displaystyle \int_0^3 at^2 + 3 \, dt = 0$

$\left[\dfrac{at^3}{3} + 3t\right]_0^3 = 0
$

$9a + 9 = 0$

$a = -1$

$v(t) = 3 - t^2$

$v(3) = 3 - 9 = -6$ , so its speed at $t = 3$ is $|-6| = 6 \, m/s$

total distance = $\displaystyle \int_0^3 |3-t^2| \, dt = \int_0^{\sqrt{3}} 3 - t^2 \, dt - \int_{\sqrt{3}}^3 3 - t^2 \, dt$

you can finish up by finding the total distance.

6. Thanks skeeter! I didn't know that you could use definite integrals.