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.