Question
A motorcyclist heading east through a small Iowa city accelerates after he passes the signpost marking the city limits. His acceleration is a constant 4.0 ms-2. At time t = 0 he is 5.0 m east of the signpost, moving east at 15 m/s.
(a) find his position and velocity at time t = 2.0 s.
(b) Where is the motorcyclist when his velocity is 25 m/s?
Answer
a) The answer to his position and velocity at time 2.0 seconds is 43 m and 23 m/s.
b) The location of the motorcyclist when his velocity is 25 m/s is 55 m.
Explanation
a)
Data: Information from the question
from kinematic equations
Distance is s1 = ut + (1/2)at2
Additionally, we can also define distance as s2 = (x-x0)/t
When we assume s1 = s2
We can see that
ut + (1/2)at2 = (x-x0)/t
Therefore, we can now make x subject of the formula
x = x0 + ut + (1/2)at2
and x0 = 5.0 m, u = 15 m/s, t = 2.0 s, a = 4.0 m/s2
x = 5 + 15 x 2 + (1/2) x 4 x 22 = 43 m
Therefore, the position of the motorcyclist at t = 2.0 s is 43 meters (m).
To find the velocity, we will use the equation v = u + at
Thus, v = 15 + 4 x 2 = 15 + 8 = 23 m/s
Therefore, the velocity of the motorcyclist at time 2.0 s is 23 meters per second (23 m/s).
b
We will use the equation v2 = u2 + 2a(x-x0)
And we will make x subject of the formula to obtain
x = x0 + (v2 – u2)/2a
By substituting the above equation with our data (remember that v = 25), we will now have
x = 5.0 + (252 – 152)/(2 x 4) = 55m
Therefore, the position of the motorcyclist at v = 25 m/s is 55 meters (m).
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