College Physics Problem 2.25

At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00\:m/s^2.
(a) How far does she travel in the next 5.00 s?
(b) What is her final velocity?
(c) Evaluate the result. Does it make sense?

Solution:

Part (a). For this part, we use the formula x=x_ot+v_ot+\frac{1}{2}at^2.

x=x_ot+v_ot+\frac{1}{2}at^2

x=0\:m+\left(9.00\:m/s\right)\left(5.00\:s\right)+\frac{1}{2}\left(-2.00\:m/s^2\right)\left(5.00\:s\right)^2

x=20\:meters

Part (b). The final velocity can be determined using the formula v=v_o+at

v=v_o+at

v=9.00\:m/s+\left(-2.00\:m/s^2\right)\left(5.00\:s\right)

v=-1\:m/s

Part (c). The result says that the runner starts at the rate of 9 m/s and decelerates at 2 m/s^2. After some time, the velocity is already negative. This does not make sense because if the velocity is negative, that means that the runner is already running backwards.

You can now buy the complete solution manual for College Physics. Just complete the google form below.

Advertisements