Distance travelled questions with answers

Questions

1. Question 1In the first T seconds

   A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$, its velocity, $v$ m/s, is given by $v = 3t^{2} - 24t + 36$. Find the total distance travelled by the particle in the first $5$ seconds.

   Show the worked solution

   1. Step 1: Does it turn back? Find when v = 0$3t^{2} - 24t + 36 = 0$$3(t - 2)(t - 6) = 0$$t = 2$ or $t = 6$
   2. Step 2: Where does it turn back during the time asked about? Then find s, with s = 0 when t = 0It turns back at $t = 2$.$s = \dfrac{3t^{3}}{3} - \dfrac{24t^{2}}{2} + 36t$$= t^{3} - 12t^{2} + 36t$
   3. Step 3: s = 0 at the start. Find s when it turns, and at the endWhen $t = 2$, $s = 32$When $t = 5$, $s = 5$
   4. Step 4: Add the distances of the two parts of the journey (each one positive)$\text{Distance} = \left|32 - 0\right| + \left|5 - 32\right|$$\text{Distance} = 32 + 27$$= 59$ m
2. Question 2In the first T seconds

   A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$, its velocity, $v$ m/s, is given by $v = 3t^{2} - 12t + 9$. Find the total distance travelled by the particle in the first $2$ seconds.

   Show the worked solution

   1. Step 1: Does it turn back? Find when v = 0$3t^{2} - 12t + 9 = 0$$3(t - 1)(t - 3) = 0$$t = 1$ or $t = 3$
   2. Step 2: Where does it turn back during the time asked about? Then find s, with s = 0 when t = 0It turns back at $t = 1$.$s = \dfrac{3t^{3}}{3} - \dfrac{12t^{2}}{2} + 9t$$= t^{3} - 6t^{2} + 9t$
   3. Step 3: s = 0 at the start. Find s when it turns, and at the endWhen $t = 1$, $s = 4$When $t = 2$, $s = 2$
   4. Step 4: Add the distances of the two parts of the journey (each one positive)$\text{Distance} = \left|4 - 0\right| + \left|2 - 4\right|$$\text{Distance} = 4 + 2$$= 6$ m
3. Question 3Between two times

   A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$, its velocity, $v$ m/s, is given by $v = t^{2} - 9t + 18$. Find the total distance travelled by the particle between $t = 2$ and $t = 5$.

   Show the worked solution

   1. Step 1: Does it turn back? Find when v = 0$t^{2} - 9t + 18 = 0$$(t - 3)(t - 6) = 0$$t = 3$ or $t = 6$
   2. Step 2: Where does it turn back during the time asked about? Then find s, with s = 0 when t = 0It turns back at $t = 3$.$s = \dfrac{t^{3}}{3} - \dfrac{9t^{2}}{2} + 18t$$= \dfrac{1}{3}t^{3} - \dfrac{9}{2}t^{2} + 18t$
   3. Step 3: Find s at the start, when it turns, and at the endWhen $t = 2$, $s = \dfrac{62}{3}$When $t = 3$, $s = \dfrac{45}{2}$When $t = 5$, $s = \dfrac{115}{6}$
   4. Step 4: Add the distances of the two parts of the journey (each one positive)$\text{Distance} = \left|\dfrac{45}{2} - \dfrac{62}{3}\right| + \left|\dfrac{115}{6} - \dfrac{45}{2}\right|$$\text{Distance} = \dfrac{11}{6} + \dfrac{10}{3}$$= \dfrac{31}{6}$ m
4. Question 4Between two times

   A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$, its velocity, $v$ m/s, is given by $v = 3t^{2} - 33t + 72$. Find the total distance travelled by the particle between $t = 2$ and $t = 7$.

   Show the worked solution

   1. Step 1: Does it turn back? Find when v = 0$3t^{2} - 33t + 72 = 0$$3(t - 3)(t - 8) = 0$$t = 3$ or $t = 8$
   2. Step 2: Where does it turn back during the time asked about? Then find s, with s = 0 when t = 0It turns back at $t = 3$.$s = \dfrac{3t^{3}}{3} - \dfrac{33t^{2}}{2} + 72t$$= t^{3} - \dfrac{33}{2}t^{2} + 72t$
   3. Step 3: Find s at the start, when it turns, and at the endWhen $t = 2$, $s = 86$When $t = 3$, $s = \dfrac{189}{2}$When $t = 7$, $s = \dfrac{77}{2}$
   4. Step 4: Add the distances of the two parts of the journey (each one positive)$\text{Distance} = \left|\dfrac{189}{2} - 86\right| + \left|\dfrac{77}{2} - \dfrac{189}{2}\right|$$\text{Distance} = \dfrac{17}{2} + 56$$= \dfrac{129}{2}$ m