Rennata Gas is driving through town at 25.0 m/s and begins to accelerate at a constant rate of -1.0m/s^2. Eventually Rennata comes to a complete stop. Use kinematic equations to calculate the distance which Rennata travels while decelerating.
vf^2=vi^2+2ad
d=Vi^2/2a=25^2/(2)=625/2 m
The distance traveled can be found by calculating the area between the line on the graph and the axes.
distance = 0.5*bh
distace = 0.5*(25.0 s)*(25.0 m/s)
distance = 313 m
To calculate the distance that Rennata travels while decelerating, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, since Rennata comes to a complete stop)
u = initial velocity (25.0 m/s)
a = acceleration (-1.0 m/s^2)
s = distance traveled
Plugging in the values, we can solve for s:
0^2 = (25.0 m/s)^2 + 2(-1.0 m/s^2)s
Simplifying the equation:
0 = 625.0 m^2/s^2 - 2s
Rearranging the equation to solve for s:
2s = 625.0 m^2/s^2
s = 312.5 m
Therefore, Rennata travels a distance of 312.5 meters while decelerating.
To calculate the distance traveled by Rennata while decelerating, we need to use the kinematic equations.
The relevant kinematic equation for this scenario is:
v^2 = u^2 + 2as
where,
v = final velocity (0 m/s, since Rennata comes to a complete stop)
u = initial velocity (25.0 m/s)
a = acceleration (-1.0 m/s^2)
s = distance traveled (unknown)
Rearranging the equation to solve for s, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values into the equation:
s = (0^2 - 25.0^2) / (2 * -1.0)
s = (-625.0) / (-2.0)
s = 312.5 meters
Therefore, Rennata travels a distance of 312.5 meters while decelerating.