#### Answer

(a) The deer is 5.0 meters from the car when the car stops.
(b) v = 21.9 m/s

#### Work Step by Step

(a) It takes 0.50 seconds to apply the brakes. We can find the distance the car moves in that time;
$d = v~t = (20~m/s)(0.50~s)$
$d = 10~m$
The car is 25 meters from the deer when the brakes are applied. We can find the distance the car moves after the brakes are applied.
$x = \frac{v^2-v_0^2}{2a}$
$x = \frac{0-(20~m/s)^2}{(2)(-10~m/s^2)}$
$x = 20~m$
The deer is 5.0 meters from the car when the car stops.
(b) The total distance the car can travel is 35 meters;
$(0.50~s)~v + \frac{v^2}{20~m/s^2}=35~m$
$v^2+10~v-700 = 0$
We can use the quadratic formula to find $v$;
$v = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$v = \frac{-(10)\pm \sqrt{(10)^2-(4)(1)(-700)}}{(2)(1)}$
$v = -31.9~m/s, 21.9~m/s$
Since the negative value is unphysical, the solution is $v = 21.9~m/s$