Question #8f27f

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Question #8f27f

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Answer 1 · 2016-09-14T07:43:02

The problem is as shown in the figure below. Momentarily both the bus and Sophia are still and are $60 m$ $60m$ apart. On looking at the bus Sophia starts running at $6 m s^{- 1}$ $6ms^-1$ towards the bus and bus drives away with a constant acceleration of $0.18 m s^{- 2}$ $0.18ms^-2$
my comp
1. Suppose Sophia catches the bus after time $t sec$ $t " sec"$
Kinematic equation for Sophia
Distance ran $= 6 t$ $=6t$
In this duration Bus moves
$s_{b u s} = u t + \frac{1}{2} a t^{2}$ $s_(bus)=ut+1/2at^2$
$\Rightarrow s_{b u s} = 0 \times t + \frac{1}{2} \times 0.18 t^{2}$ $=>s_(bus)=0xxt+1/2 xx0.18t^2$
$\Rightarrow s_{b u s} = 0.09 t^{2}$ $=>s_(bus)=0.09t^2$

To catch the bus, distance run by Sophia $6 t = 60 + 0.09 t^{2}$ $6t=60+0.09t^2$
Rewriting we obtain
$0.09 t^{2} - 6 t + 60 = 0$ $0.09t^2-6t+60=0$
Multiplying both sides with $\frac{100}{3}$ $100/3$ we get
$3 t^{2} - 200 t + 2000 = 0$ $3t^2-200t+2000=0$
using the formula to find roots of a quadratic
$t = \frac{200 \pm \sqrt{40000 - 4 \times 3 \times 2000}}{2 \times 3}$ $t=(200+-sqrt(40000-4xx3xx2000))/(2xx3)$
$\Rightarrow t = \frac{200 \pm \sqrt{16000}}{6}$ $=>t=(200+-sqrt(16000))/(6)$
$\Rightarrow t = \frac{200 \pm 126.5}{6}$ $=>t=(200+-126.5)/(6)$

Selecting $- v e$ $-ve$ sign as that relates to time when Sophia catches the bus
$t = 12.25 s$ $t=12.25s$ , rounded to two decimal places.

Distance run by Sophia in this time $= 12.25 \times 6 = 49.5 m$ $=12.25xx6=49.5m$ , rounded to one decimal place.
2. There is no change in the kinematic equation for the bus. But for Sophia we have
Distance ran $= 4 t$ $=4t$

To catch the bus, distance run by Sophia $4 t = 60 + 0.09 t^{2}$ $4t=60+0.09t^2$
Rewriting we obtain
$0.09 t^{2} - 4 t + 60 = 0$ $0.09t^2-4t+60=0$

Multiplying both sides with $100$ $100$ we get
$9 t^{2} - 400 t + 6000 = 0$ $9t^2-400t+6000=0$

Now this time
$t = \frac{400 \pm \sqrt{160000 - 4 \times 9 \times 6000}}{2 \times 9}$ $t=(400+-sqrt(160000-4xx9xx6000))/(2xx9)$
$\Rightarrow t = \frac{400 \pm \sqrt{- 56000}}{2 \times 9}$ $=>t=(400+-sqrt(-56000))/(2xx9)$

We see that discriminant is $- v e$ $-ve$ and square root of this number is imaginary. As such the real roots don't exist. Sophia will never be able to catch the bus.
3. Suppose Sophia needs to run at a velocity of $v m s^{- 1}$ $v" "ms^-1$ in order to catch the bus.
There is no change in the kinematic equation for the bus. But for Sophia we have
Distance ran $= v t$ $=vt$
To catch the bus, distance run by Sophia $v t = 60 + 0.09 t^{2}$ $vt=60+0.09t^2$
Rewriting we obtain
$0.09 t^{2} - v t + 60 = 0$ $0.09t^2-vt+60=0$

Multiplying both sides with $100$ $100$ we get
$9 t^{2} - 100 v t + 6000 = 0$ $9t^2-100vt+6000=0$
Using the formula for roots of a quadratic
$t = \frac{100 v \pm \sqrt{10000 v^{2} - 4 \times 9 \times 6000}}{2 \times 9}$ $t=(100v+-sqrt(10000v^2-4xx9xx6000))/(2xx9)$
For real roots and with minimum velocity required for Sophia to run the discriminant must be set to be $= 0$ $=0$ . We have
$10000 v^{2} - 4 \times 9 \times 6000 = 0$ $10000v^2-4xx9xx6000=0$
Solving for $v$ $v$
$10000 v^{2} = 216000$ $10000v^2=216000$
Ignoring the $- v e$ $-ve$ root
$v \approx 4.65 m s^{- 1}$ $v~~4.65ms^-1$ , rounded to two decimal places.
$t = \frac{4.65 \times 100}{18}$ $t=(4.65xx100)/18$
$t = \frac{465}{18}$ $t=465/18$ (discriminant being zero, roots are $= - \frac{b}{2 a}$ $=-b/(2a)$ )
$t = 25.8 s$ $t=25.8s$ , rounded to one decimal place.

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