A motorcycle is traveling at 20 m/s when it enters the test section of a circular track with a 1000-m radius. Once in the test section, the cycle slows down and its deceleration along the track varies with its position, s, by the relation a(subscript t) = -2.10s^2 (m/s^2). Determine
1.) The distance traveled within the test section until it comes to a stop.
2.) The magnitude of its acceleration at s = 5 m.
To determine the distance traveled within the test section until the motorcycle comes to a stop, we need to find the position at which its velocity becomes zero.
1.) Distance traveled until the motorcycle comes to a stop:
To find the position at which the motorcycle comes to a stop (s = s_stop), we need to find the time it takes for the velocity to become zero. We can use the relationship between acceleration, velocity, and position:
v = u + at
where:
v is the final velocity (which we want to be zero),
u is the initial velocity (20 m/s),
a is the acceleration as a function of position (-2.10s^2 m/s^2),
t is the time.
Rearranging the equation, we have:
t = (v - u) / a
Since v equals zero (as the motorcycle comes to a stop), we have:
t = (-u) / a
Substituting the given values:
u = 20 m/s
a = -2.10s^2 m/s^2
t = (-20 m/s) / (-2.10s^2 m/s^2)
t = 9.52 seconds (rounded to two decimal places)
Now, to find the distance traveled (d_stop) until the motorcycle comes to a stop:
Using the equation of motion:
s = ut + (1/2)at^2
where:
s is the distance,
u is the initial velocity (20 m/s),
a is the acceleration as a function of position (-2.10s^2 m/s^2),
t is the time (9.52 seconds).
Substituting the values:
s_stop = (20 m/s)(9.52 s) + (1/2)(-2.10 s^2 m/s^2)(9.52 s)^2
s_stop = 190.40 m + (-99.06 m) ≈ 91.34 meters
Therefore, the motorcycle will travel approximately 91.34 meters before coming to a stop within the test section.
2.) Magnitude of acceleration at s = 5 m:
To find the magnitude of acceleration at s = 5 m, we can directly substitute s = 5 m into the given relation between acceleration and position, a(subscript t) = -2.10s^2 m/s^2:
a(5) = -2.10(5)^2 m/s^2
a(5) = -2.10(25) m/s^2
a(5) = -52.50 m/s^2
Therefore, the magnitude of the acceleration at s = 5 m is 52.50 m/s^2.