It is observed that rider A closes the 48 gap in a time of 6 seconds (a=ft/s^2). a) What speed was rider A moving while located 48 feet behind rider B? b) As timed from when rider A was 48 feet behind rider B, how many seconds elapse before rider A matches rider B's speed? (rider B has a constant speed of 54ft/s.)
To solve these questions, we can use the equations of motion.
a) To find the speed of rider A when they were 48 feet behind rider B, we need to determine their initial speed.
We know the acceleration of rider A (a) is given as 48 ft/s^2. Since we are assuming that the acceleration is constant, we can use the following equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Initially, rider A was 48 feet behind rider B, so the initial velocity is 0 ft/s (since they were stationary). We need to find the final velocity when they close the 48-foot gap in 6 seconds.
Using the formula, we can substitute the values:
v = 0 + (48 ft/s^2)(6 s)
v = 0 + 288 ft/s
v = 288 ft/s
Therefore, the speed of rider A when they were located 48 feet behind rider B was 288 ft/s.
b) To find out how many seconds elapse before rider A matches rider B's speed, we need to determine the time it takes for rider A to reach a speed of 54 ft/s, which is the speed of rider B.
Again, we can use the same equation:
v = u + at
But this time, we know the final velocity (v) is 54 ft/s and the initial velocity (u) is 0 ft/s (as rider A starts from rest). The acceleration (a) remains the same at 48 ft/s^2.
Now we need to solve for time (t):
54 ft/s = 0 + (48 ft/s^2) * t
Rearranging the equation, we get:
48t = 54 ft/s
t = 54 ft/s / 48 ft/s^2
t = 1.125 s
Therefore, it takes approximately 1.125 seconds for rider A to match rider B's speed after initially being 48 feet behind rider B.