A student is running at her top speed of 5.5m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 42.0m from the bus, it starts to pull away, moving with a constant acceleration of 0.179m/s2 .
For how much time does the student have to run at 5.5m/s before she overtakes the bus?
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Let us say the student catch the bus t seconds after the bus start travel.
In this time student runs at 5.5m/s constant speed and bus accelerates at 0.179m/s^2.
When she catches the bus let us say bus has gone xm from its initial position. That means the student has run (42+x) to catch the bus.
Using equations of motion;
`S = ut+1/2at^2`
`x = 0+1/2xx0.179t^2----(1)`
`42+x = 5.5t ---(2)`
`-42 = 0.179/2t^2-5.5t`
`-84 = 0.179t^2-11t`
`0 = 0.179t^2-11t+84`
Solving the above quadratic equation give you;
`t = (11+-sqrt((-11)^2-4xx0.179xx84))/(2xx0.179)`
t = 52.51 OR t = 8.93
There are two answers. Both are valid since they are positive. We must be tactical here. Initially student has more speed than bus because it start from rest. So the student can easiliy overtake the bus and go beyond. But since the bus is accelerating at some time it will overtake the student again. So the required answer is t = 8.93s.
A Kinematic Equation for Distance is:
`d = v_(i)(t) + 1/2at^2`
Equation for student:
`v_(i) = 5.5 m/s`
`d= 5.5(t) + 1/1 (0) t^2`
`d = 5.5t`
Equation for bus:
The student must run 42.0 m to catch up to the bus, so the student's distance equals the bus' distance plus 42.0:
`5.5t=1/2(.176)t^2 + 42`
`0=.0895t^2 - 5.5t + 42`
Solving the quadratic:
`t = (5.5 +- sqrt(5.5^2 - 4(.0895)(42)))/(2(.0895))`
We choose the first answer, because that is when the student and the bus meet for the first time. The second number represents the time when the bus passes the student, some time later.
It takes the student 8.9 seconds to catch up to the bus.
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