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) + (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) + (1)/(2) a(8.3\:s)^(2)\\ 250\:m = (60.0\:m/s)(8.3\:s) + (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)`.