91Ó°ÊÓ

• 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-

$t=\frac{V_0Â\pm \sqrt{V_0^2-2a_b{x}_{bo}}}{a_b}$

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,

$d_t=V_{min}×t$

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" $3.668\mathrm{m}/\mathrm{s}$