Final answer:
To find the position and velocity of the motorcyclist at t=2.0 s, we use the kinematic equations for constant acceleration. The final velocity is 23 m/s, and the total distance from the start is 43 m east.
Step-by-step explanation:
The student is asking to find the position and velocity of a motorcyclist who accelerates at a constant acceleration from a given initial velocity and position. The motorcyclist starts 5 meters east of the city-limits signpost with an initial velocity of 15 m/s, and accelerates at 4.0 m/s2 eastward.
To determine the motorcyclist's position and velocity at t=2.0 s, we can use the kinematic equations for uniformly accelerated motion.
The final velocity (v) can be found using the equation:
v = u + at
Where u is the initial velocity, a is the acceleration, and t is the time. Thus, we get:
v = 15 m/s + (4.0 m/s2)(2.0 s) = 23 m/s
The position (s) can be found using the equation:
s = ut + 1/2 at2
Where s is the displacement from the starting point. Hence:
s = (15 m/s)(2.0 s) + 1/2 (4.0 m/s2)(2.0 s)2 = 30 m + 8 m = 38 m
The final position of the motorcyclist at t=2.0 s is 5.0 m + 38 m = 43 m east of the city-limits signpost.