# Math Help - Two Distance-velocity related questions

1. ## Two Distance-velocity related questions

1)A particle move in a straight line so that its distance, s metres, from a fixed point is given by s=t^3(2-t)^2, where t is the time in seconds aftre passing through the fixed point.

Find (i) the values of t when the particle is at rest
(ii)the greatest distance travelled by the particle.

For 1)(i)i think the velocity is 0 when the paricle is at rest. So i ds/dt=0
However i could not get the values right. my ds/dt is 12t^2-32t^3+5t^4. I checked the ans but it is wrong. can anyone help me please?

For1)(ii)i think the greatest distance travelled is when velocity is at =0 Is it correct?

Another question:A particle P starts from point A and moves in a straight line so that its displacement s in metres, from a fixed point O, t seconds after leaving A is given by s=5+t(t-5)^2

Find the total distance travelled from t=0 to t=5.

2. Originally Posted by helloying
1)A particle move in a straight line so that its distance, s metres, from a fixed point is given by s=t^3(2-t)^2, where t is the time in seconds aftre passing through the fixed point.

Find (i) the values of t when the particle is at rest
(ii)the greatest distance travelled by the particle.

For 1)(i)i think the velocity is 0 when the paricle is at rest. So i ds/dt=0
However i could not get the values right. my ds/dt is 12t^2-32t^3+5t^4. I checked the ans but it is wrong. can anyone help me please? Mr F says: Please show the details of how you got your answer. Using the product rule, I get ds/dt = t^2 (2 - t) (6 - 4t).

For1)(ii)i think the greatest distance travelled is when velocity is at =0 Is it correct? Mr F says: There is no finite answer if the question is as you've worded it.

Another question:A particle P starts from point A and moves in a straight line so that its displacement s in metres, from a fixed point O, t seconds after leaving A is given by s=5+t(t-5)^2

Find the total distance travelled from t=0 to t=5.
I suggest you draw a graph of s versus t and use it to add up the distances.

3. Hello, helloying!

1) A particle move in a straight line so that its distance, s metres,
from a fixed point is given by: $s\:=\:t^3(2-t)^2$,
where $t$ is the time in seconds after passing through the fixed point.

(a) Find the values of $t$ when the particle is at rest.
You are right . . . We want: . $s' \:=\:0$

Product Rule: . $s' \;=\;t^3\cdot2(2-t)(-1) + 3t^2(2-r)^2 \:=\:-2t^3(2-t) + 3t^2(2-t)^2 \:=\:0$

Factor: . $t^2(2-t)\bigg[-2t + 3(2-t)\bigg]\:=\:0 \quad\Rightarrow\quad t^2(2-t)(6-5t) \:=\:0$

Therefore: . $t \;=\;0,\:\tfrac{6}{5},\:2$

(b) Find the greatest distance travelled by the particle.
The problem should be more specific.

In theory, the particle moves forever.
So that at $t = 100\!:\;\;s \:=\:9,\!605,\!000,\!000$ and is moving away from the fixed point.

If we are concerned with only the first two seconds, we have an answer.

We have: . $\begin{array}{ccc}s(0) &=& 0 \\ s\left(\tfrac{6}{5}\right) &=& 1.10592 \\ s(2) &=& 0 \end{array}$

It starts at the fixed point.
In the next 1.2 seconds, it moves 1.10592 units to the right.
In the next 0.8 seconds, it moves 1.10592 units back to the fixed point.

The "greatest distance" is in there somewhere.

2) A particle $P$ starts from point $A$ and moves in a straight line
so that its displacement s in metres, from a fixed point $O,\:t$ seconds after leaving $A$
is given by: . $s\:=\:5+t(t-5)^2$

Find the total distance travelled from $t=0$ to $t=5$.

We have: . $s(t) \:=\:t^3 - 10t^2 + 25t + 5$

Then: . $s'(t) \:=\:3t^2 - 20t + 25 \:=\:0 \quad\Rightarrow\quad (3t-5)(t-5) \:=\:0$

Hence: . $t\:=\:\tfrac{5}{3},\:5$

We have: . $\begin{array}{ccccc}s(0) &=& 5 + 0(0-5)^2 &=& 5 \\ \\[-4mm] s(\frac{5}{3}) &=& 5 + \frac{5}{3}(\frac{5}{3}-5)^2 &=& \frac{635}{27} \\ \\[-4mm] s(5) &=& 5 + 5(5-5)^2 &=& 5\end{array}$

At $t = 0$, particle $P$ is 5 units to the right of fixed point $O.$

At $t = \tfrac{5}{3},\:P$ is $23\tfrac{14}{27}$ units to the right of $O.$
. . It has moved: . $23\tfrac{14}{27} - 5 \:=\:18\tfrac{14}{27}$ units to the right.

At $t = 5,\:P$ is 5 units to the right of $O.$
. . It has moved: . $23\tfrac{14}{27}-5 \:=\:18\tfrac{14}{27}$ units to the left.

The total distance is: . $18\tfrac{14}{27} + 18\tfrac{14}{27} \:=\:37\tfrac{1}{27}$ units.