# College Physics Problem 2.25

#### 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.

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