Here we will look at a more complex problem for which we will use all three of o
ID: 1405936 • Letter: H
Question
Here we will look at a more complex problem for which we will use all three of our primary constant acceleration equations. A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His acceleration is constant: ax=4.0m/s2. At time t=0, he is 5.0 m east of the signpost and has a velocity of v0x=15m/s. (a) Find his position and velocity at time t=2.0s. (b) Where is the motorcyclist when his velocity is 25 m/s?
a.) If the acceleration is 2.0 m/s2 instead of 4.0 m/s2, where is the cyclist, 5.0 s after he passes the signpost if x0=5.0m and v0x=15m/s?
Express your answer in meters to the nearest integer.
b.) How fast is he moving, 5.0 s after he passes the signpost if x0=5.0m and v0x=15m/s?
Express your answer in meters per second to the nearest integer.
Explanation / Answer
given,
acceleration = 4 m/s^2
initial velocity = 15 m/s
initial position = 5 m
by second equation of motion
s = ut + 0.5 * s * t^2
s = 15 * 2 + 0.5 * 4 * 2^2
s = 38 m
position after 2 sec = 38 + 5
position after 2 sec = 43 m
by first equation of motion
v = u + at
v = 15 + 4 * 2
v = 23 m/s
velocity after 2 sec = 23 m/s
if final velocity is 25 m/s then,
by third equation of motion
v^2 = u^2 + 2as
25^2 = 15^2 + 2 * 4 * s
s = 50 m
position when velocity is 25 m/s = 50 + 5
position when velocity is 25 m/s = 55 m
if acceleration is 2 m/s^2 instead of 4 m/s^2 then
s = ut + 0.5 * at^2
s = 15 * 5 + 0.5 * 2 * 5^2
s = 100 m
position after 5 sec = 100 + 5
position after 5 sec = 105 m
v = u + at
v = 15 + 2 * 5
v = 25 m/s
velocity after 5 sec = 25 m/s