At time t = 0 s, automobile A is traveling at a speed of 18.6 m/s along a straight road and is picking up speed with an acceleration that has a magnitude of 3.70 m/s2 as in part a of the figure. At t = 0 s, automobile B is traveling at a speed of 18.6 m/s in uniform circular motion as it negotiates a turn (part b of the figure). It has a centripetal acceleration whose magnitude is also 3.70 m/s2. Determine the speed of (a) automobile A and (b) automobile B, when t = 2.0 s.
To determine the speed of automobile A and automobile B when t = 2.0 s, we need to use the equations of motion.
a) For automobile A:
We know that the initial velocity (u) is 18.6 m/s and the acceleration (a) is 3.70 m/s^2. We want to find the speed (v) when t = 2.0 s.
The equation we can use is:
v = u + at
Substituting the given values, we have:
v = 18.6 m/s + (3.70 m/s^2)(2.0 s)
Calculating the right side:
v = 18.6 m/s + 7.40 m/s
v = 26.0 m/s
Therefore, the speed of automobile A when t = 2.0 s is 26.0 m/s.
b) For automobile B:
We know that the centripetal acceleration (a) is 3.70 m/s^2 and we want to find the speed (v) when t = 2.0 s.
The equation relating centripetal acceleration and speed is:
a = v^2 / r
Where r is the radius of the circular path.
Since the radius is not given, we need more information to determine the speed of automobile B when t = 2.0 s.