Question

When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters from its starting position in 8.3 s. Calculate the acceleration.

Answer 1

Remember your kinematic equations for constant acceleration. One of the equations is $x_{f} = x_{i} + v_{i}(t) + \frac{1}{2} at^{2}$, where $x_{f}$ = final position, $x_{i}$ = initial position, $v_{i}$ = initial velocity, t = time, and a = acceleration. 

Your initial position is where you initially were before you braked. That means $x_{i}$ = 100m. You final position is where you ended up after t seconds passed, so $x_{f}$ = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was $v_{i}$ = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
$350\:m = 100\:m + (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\ 250\:m = (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\ 250\:m = 498\:m +34.445\:s^{2}(a)\\ -248\:m = 34.445\:s^{2}(a)\\ a \approx -7.2 \: m/s^{2}$

Your acceleration is approximately $-7.2 \: m/s^{2}$.