Answer
(a) The student hops on the bus at time $t=9.55s$ after running a distance of $47.8m$.
(b) The speed of the bus at time $t=9.55s$ is $1.62\frac{m}{s}$.
(c) The sketch of the $x-t$ graph for both the student and the bus is shown below.
(d) The speed of the bus at time $t=49.3s$ is $8.38\frac{m}{s}$.
(e) No, the student will not catch the bus if she runs at $3.5\frac{m}{s}$.
(f) The minimum speed that the student must have to just catch up with the bus is $3.69\frac{m}{s}$.
The student must run a distance of $77.3m$ for time $t\approx21.0s$ before hoping on the bus.
Work Step by Step
We take the origin of our system to be at the initial position of the student when the bus is $40.0m$ away from her.
At $t=0$ the student is at the origin and the bus is $40.0m$ away from her at rest with an acceleration of $0.170\frac{m}{s^2}$.
Since the motions are of constant acceleration, we can use the constant-acceleration kinematics equation.
(a) The position of the student $x_{s}$ and the bus $x_{b}$ as a function of time is given by
$x_{s}=5.0\frac{m}{s}t$ and $x_{b}=40.0m+\frac{0.170\frac{m}{s^2}}{2}t^2$
For the student to overtake the bus, the student and the bus should be at the same position at the same time i.e,
$x_{s}=x_{b}$
$=>5.0\frac{m}{s}t=40.0m+\frac{0.170\frac{m}{s^2}}{2}t^2$
$=>\frac{0.170\frac{m}{s^2}}{2}t^2-5.0\frac{m}{s}t+40.0m=0$
Using the quadratric formula, we get
$t=\frac{5.0+-\sqrt ((-5.0)^2-4\times\frac{0.170}{2}\times40.0)}{2\times\frac{0.170}{2}}=49.3s$ or $9.55s$
Now, the distance travelled by her in time $t=9.55s$ is
$x_{s}=5.0\frac{m}{s}\times9.55s=47.8m$
The student hops on the bus at time $t=9.55s$ after running a distance of $47.8m$.
(b) The speed of the bus at time $t=9.55s$ is given by
$v_{b}=0.170\frac{m}{s^2}\times9.55s=1.62\frac{m}{s}$
(c) The sketch of the $x-t$ graph for both the student and the bus is shown below.
(d) Suppose that the student did not hop on the bus at time $t=9.55s$ and continued the motion, the student will overtake the bus. Then at a later time $t=49.3s$ the bus catches up to her and overtakes her.
The speed of the bus at time $t=49.3s$ is given by
$v_{b}=0.170\frac{m}{s^2}\times49.3s=8.38\frac{m}{s}$
(e) Suppose the student runs at $3.5\frac{m}{s}$, then
$\frac{0.170\frac{m}{s^2}}{2}t^2-3.5\frac{m}{s}t+40.0m=0$
Using quadratric formula, we get
$t=\frac{3.5+-\sqrt ((-3.5)^2-4\times\frac{0.170}{2}\times40.0)}{2\times\frac{0.170}{2}}=$imaginary solutions
Since, $(-3.5)^2\lt2\times0.170\times40.0$, there are no real solutions.
So, no the student will not catch the bus if she runs at $3.5\frac{m}{s}$.
(f) Let $v_{s}$ be the velocity of the student.
Then, $\frac{0.170\frac{m}{s^2}}{2}t^2-v_{s}t+40.0m=0$
Using quadratic formula, we get
$t=\frac{v_{s}+-\sqrt ((-v_{s})^2-4\times\frac{0.170}{2}\times40.0)}{2\times\frac{0.170}{2}}$
For $t$ to be real , $(-v_{s})^2\gt4\times\frac{0.170}{2}\times40.0$
$v_{s}=\sqrt (2\times0.170\times40.0)=3.69\frac{m}{s}$
The minimum speed that the student must have to just catch up with the bus is $3.69\frac{m}{s}$.
Now, $t=\frac{3.69+-\sqrt ((-3.69)^2-4\times\frac{0.170}{2}\times40.0)}{2\times\frac{0.170}{2}}=22.45s$ or $20.96$
The distance travelled by the student at time $t=20.96s\approx21.0s$ is given by
$x_{s}=3.69\frac{m}{s}\times20.96s=77.3m$
The student must run a distance of $77.3m$ for time $t\approx21.0s$ before hoping on the bus.
