An object is initially moving in the positive direction of an x-axis with a speed of 2.7 m/s. If the constant acceleration of the object is -0.12 m/s2, what distance will it have traveled when it turns around?
What must be the radius of an unbanked (flat) curve so that a car can safely travel at a maximum speed of 39 m/s? The coefficient of static friction is 0.70.
v = Vi + a t
v = 0 when it stops
0 = 2.7 - .12 t
y = 2.25 seconds to stop
d = Vi t - (1/2) a t^2
= 2.7 * 2.25 - .5 * .12 * 2.25^2
To determine the distance the object will have traveled when it turns around, we need to find the time it takes for the object to come to a stop. We can use the equation of motion:
v = u + at
where:
v = final velocity (0 m/s, when the object comes to a stop)
u = initial velocity (2.7 m/s, the speed at which the object is initially moving)
a = acceleration (-0.12 m/s^2)
Rearranging the equation, we have:
t = (v - u) / a
Substituting the values into the equation:
t = (0 - 2.7) / -0.12
Simplifying the equation:
t = 22.5 seconds
Next, to find the distance traveled, we can use the equation of motion:
s = ut + (1/2)at^2
where:
s = distance traveled
u = initial velocity (2.7 m/s)
t = time (22.5 seconds)
a = acceleration (-0.12 m/s^2)
Substituting the values into the equation:
s = (2.7)(22.5) + (1/2)(-0.12)(22.5)^2
Simplifying the equation:
s = 60.75 - 15.1875
s ≈ 45.56 meters
Therefore, the object will have traveled approximately 45.56 meters when it turns around.