Question

A student is running at her top speed of 5.1m/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 0.170m/s2 .

For how much time does the student have to run at 5.1m/s before she overtakes the bus?

For what distance does the student have to run at 5.1m/s before she overtakes the bus?

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

If the student's top speed is 3.00m/s , will she catch the bus?

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

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

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

Answer 1

1. 5.1t = 42+0.5*0.17t^2

=> t = 9.85 sec

2. distance = 5.1t = 50.235 m

3. velocity = at = 0.17*t = 1.675 m/s

4. minimum speed the student must have to just catch up with the bus-

vt = 42+0.5*0.17t^2

=> 0.085t^2 - vt +42 = 0

for minimum speed-

b^2 - 4ac = 0

=> v^2 - 4(0.085*42) =0

=> v = 3.78 m/sec

so,

If the student's top speed is 3.00m/s , she will not catch the bus

5

minimum speed the student must have to just catch up with the bus-

vt = 42+0.5*0.17t^2

=> 0.085t^2 - vt +42 = 0

for minimum speed-

b^2 - 4ac = 0

=> v^2 - 4(0.085*42) =0

=> v = 3.78 m/sec

6.

3.78t = 42+0.5*0.17t^2

so, t = 22.23 sec

7 distance = vt = 3.78*22.23= 84 m