To find the average velocity during each 2-second interval, we first need to determine the acceleration of the bus.
Let's denote the initial velocity of the bus as u, the acceleration as a, the distance from the station as s, and the time since passing the 92-ft point as t.
From the information given:
- At t=2 s, s=140 ft
- At t=4 s, s=198 ft
- At t=6 s, s=266 ft
Using the equation of motion:
s = ut + (1/2)at^2
For the period between 2 s and 4 s:
(98) = 2u + 2a
(1) u + a = 49
For the period between 4 s and 6 s:
(68) = 4u + 4a
(2) u + a = 17
Solving these two equations, we get:
u = 15 ft/s and a = 34 ft/s^2
Now, let's find the average velocity during each 2-sec interval after passing the 92-ft point:
- From 2 s to 4 s:
Initial velocity, u = 15 ft/s
Final velocity, v = u + at = 15 + 34*2 = 83 ft/s
Average velocity = (v+u)/2 = (83+15)/2 = 49 ft/s
- From 4 s to 6 s:
Initial velocity, u = 83 ft/s
Final velocity, v = u + at = 83 + 34*2 = 151 ft/s
Average velocity = (v+u)/2 = (151+83)/2 = 117 ft/s
Therefore, the average velocity of the bus during each 2-sec interval after passing the 92-ft point is 49 ft/s and 117 ft/s, respectively.
A bus starts from rest at a station and runs with constant acceleration. The bus is 140 ft from the station 2 sec after passing a point which is 92 ft from the station. Four seconds after passing the 92-ft point, it is 198 ft from the station, and 6 sec after passing this point, it is 266 ft from the station.
a) What is the average velocity of the bus during each 2-sec interval after passing the 92-ft point?
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