A student is running at her top speed of 5.0 m/s to catch a bus, which is stopped at the bus stop. When the student is still 40.0 m from the bus, it starts to pull away, moving with a constant acceleration of 0.170 m/s². (a) For how much time and what distance does the student have to run at 5.0 m/s before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an x-tgraph for both the student and the bus. Take x= 0 at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student’s top speed is 3.5 m/swill she catch the bus? (f) What is the minimumspeed the student must have to just catch up with the bus?For what time and what distance does she have to run in that case?

• The student’s constant speed is, ${\mathrm{v}}_0=5.0\mathrm{m}/\mathrm{s}$
• The initial position of the bus, ${\mathrm{x}}_0=40.0\mathrm{m}$
• The constant acceleration of the bus,$\mathrm{a}=0.170\mathrm{m}/{\mathrm{s}}^2$

When an item moves, it has a certain velocity. It is a vector quantity that is it has magnitude and direction.

Speed is the pace at which an item travels on a route.

a)

The position of the student as function of time is,

$x_1=v_{0}t$… (i)

Here, $x_1$is the initial displacement of the bus, $v_0$is the initial speed of the bus and t is the time.

The position of the bus as function of time is,

$x_2=x_0+\left(\frac{1}{2}a{t}^2\right)$… (ii)

Here, $x^2$is the displacement at the second point and a is the acceleration

Equating equation (i) and equation (ii) we get,

$$\begin{gathered} v_{0}t=x_0+\frac{1}{2}a{t}^2 \\ \frac{1}{2}a{t}^2-v_{0}t+x_0=0 \end{gathered}$$

Substituting the values in the above equation,

$t=\frac{1}{a}\left(v_0+\sqrt{v_0^2-2a{x}_0}\right)$… (iii)

Substitute values in equation (iii) we get,

$$\begin{gathered} t=\frac{1}{0.170m/s}\left(5.0m/s+\sqrt{{\left(5.0m/s\right)}^2-\left(0.170m/s^2\right)\left(40.0m\right)}\right) \\ =9.55sand49.27s \end{gathered}$$

The student overtakes the bus at$t=9.55s$but as she decided to run, she overtook the bus. After the time $t=49.27s$, the bus is going to catch her again.

The first time the student sees the bus, she is likely to board it. The speed at this time is,

$d=v_{0}t$

Here, d is the distance and t are the time

Substituting the values in the above equation,

$$\begin{gathered} d=\left(5m/s\right)\left(9.55s\right) \\ =47.75m \end{gathered}$$

Thus, the speed at which the student hops on bus is, $47.75m$at $t=9.55s$.

b)

After some time, the student passes the bus maintaining constant speed. At this time the bus speed can be evaluated as,

${\mathrm{V}}_{\mathrm{b}}=\mathrm{at}$

Here, ${\mathrm{V}}_{\mathrm{b}}$is the speed of the bus.

Substituting the values in the above equation,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{b}}=(0.170\mathrm{m}/{\mathrm{s}}^2)(9.55\mathrm{s}) \\ =1.62\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the bus is travelling at$1.62\mathrm{m}/\mathrm{s}$

c)

the x-t graph has been drawn below-

d)

At this time the bus speed can be evaluated as,

${\mathrm{V}}_{\mathrm{b}}=\mathrm{at}$

Here, ${\mathrm{V}}_{\mathrm{b}}$is the speed of the bus.

Substituting the values in the above equation,

role="math" localid="1655812791365" $$\begin{gathered} {\mathrm{V}}_{\mathrm{b}}=(0.170\mathrm{m}/{\mathrm{s}}^2)(49.27\mathrm{s}) \\ =8.38\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the bus is travelling at $8.38\mathrm{m}/\mathrm{s}$

e)

The equation of time at this point can be calculated as-

$t=\frac{V_0\pm \sqrt{V_0^2-2a_b{x}_{bo}}}{a_b}$

Here, the root part contains a negative value, then the solution become imaginary

No, she won’tbe able to catch the bus.

As,${\mathrm{V}}_0^2<2{\mathrm{a}}_{\mathrm{b}}{\mathrm{x}}_0$

The roots of the above equation are imaginary. When the student runs at$3.5\mathrm{m}/\mathrm{s}$

f)

The minimum speed at student will catch the bus is,

${\mathrm{V}}_0^2<2{\mathrm{a}}_{\mathrm{b}}{\mathrm{x}}_{\mathrm{b}0}$

The minimum speed of the student to catch bus will be,

${\mathrm{V}}_{\min }=\sqrt{2{\mathrm{a}}_{\mathrm{b}}{\mathrm{x}}_{\mathrm{b}0}}$

Here, ${\mathrm{V}}_{\min }$is the minimum velocity,${\mathrm{a}}_{\mathrm{b}}$is the acceleration and ${\mathrm{x}}_{\mathrm{b}0}$is the distance covered

Substituting the values in the above equation,

$$\begin{gathered} {\mathrm{V}}_{\min }=\sqrt{2(0.170\mathrm{m}/{\mathrm{s}}^2)(40\mathrm{m}/\mathrm{s})} \\ =3.688\mathrm{m}/\mathrm{s} \end{gathered}$$

The time she would be running is,

$\mathrm{t}=\frac{{\mathrm{V}}_{\min }}{{\mathrm{a}}_{\mathrm{b}}}$

Substituting the values in the above equation,

$$\begin{gathered} \mathrm{t}=\frac{3.68\mathrm{m}/\mathrm{s}}{0.170\mathrm{m}/{\mathrm{s}}^2} \\ =21.7\mathrm{s} \end{gathered}$$

The distance covered by the student is,

$d_t=V_{min}\times t$

Here, ${\mathrm{d}}_{\mathrm{t}}$is the distance covered by the student

Substituting the values in the above equation,

$$\begin{gathered} {\mathrm{d}}_{\mathrm{t}}=(3.688\mathrm{m}/\mathrm{s})(21.7\mathrm{s}) \\ =80.0\mathrm{m} \end{gathered}$$

Thus, when the student runs at , role="math" localid="1655811666593" $3.688\mathrm{m}/\mathrm{s}$the lines will intersect at one point, $\mathrm{x}=80\mathrm{m}$. The minimum speed of student to catch the bus is, role="math" localid="1655811519730" $3.668\mathrm{m}/\mathrm{s}$