A student is running at her top speed of 4.5 m / s to catch a bus, which is stop

Question

A student is running at her top speed of 4.5m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 38.2m from the bus, it starts to pull away, moving with a constant acceleration of 0.167m/s2 .

Part A

For how much time does the student have to run at 4.5m/s before she overtakes the bus?

Part B

For what distance does the student have to run at 4.5m/s before she overtakes the bus?

Part C

When she reaches the bus, how fast is the bus traveling?

Part D

If the student's top speed is 2.50m/s , will she catch the bus?

Part E

What is the minimum speed the student must have to just catch up with the bus?

Part F

For what time does she have to run in that case?

Part G

For what distance does she have to run in that case?

Explanation / Answer

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A student is running at her top speed of 5.4 to catch a bus, which is stopped at the bus stop. When the student is still a distance 40.6 from the bus, it starts to pull away, moving with a constant acceleration of 0.179 .
a)For how much time does the student have to run at 5.4 before she overtakes the bus?
b)For what distance does the student have to run at 5.4 before she overtakes the bus?
c)When she reaches the bus, how fast is the bus traveling?
d)If the student's top speed is 3.00 , will she catch the bus?
e)What is the minimum speed the student must have to just catch up with the bus?
f)For what time does she have to run in that case?
g)For what distance does she have to run in that case?

Initial distance between the student and bus = 40.6 m

Acceleration of the bus = 0.179 m/s2

Distance travelled by student = Distance travelled by bus + 40.6 m

(a)

So 5.4t = 1/2 x 0.179 x t2 + 40.6

which gives 0.0895 t2 - 5.4t + 40.6 = 0

Solving the quadratic eqn, we get t = 51.53 sec or t = 8.8 sec

So this means, the student catches the bus at 8.8 sec. (The second solution 51.53 sec is when the student runs past the bus, and the bus again reaches the student due to its acceleration)

(b)

Distance travelled by student = vt = 5.4t = 5.4 x 8.8 = 47.52 m

(c)

Velocity of bus at t = 8.8 sec = u + at = 0 + 0.179(8.8) = 1.575 m/s

(d)

If speed is 3, then we have the equation as 0.0895t2 - 3t + 40.6 = 0

The discriminant of this eqn is ((-3)2 - (4x0.0895x40.6) = -5.5348 < 0

So there are no real roots. This means the student can't catch the bus when speed is 3m/s

(e)

Min velocity = Value of v at which the equation 0.0895t2 - vt + 40.6 = 0 has real roots. i.e. Discriminant >= 0

v2 - (4x0.0895x40.6) >= 0 which means v >= 3.81246 m/s

So min velocity required = 3.81246 m/s

(f)

If v = 3.81246 m/s, she needs to run for time t where 0.0895t2 - 3.81246t + 40.6 = 0

Solving, we get t = 21.34 sec or 21.25 sec ... So solution can be considered as 21.3 sec

(g)

Distance the student travels = vt = 3.81246 x 21.3 = 81.2 m   

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