Question

A car traveling at 4545 ft/sec decelerates at a constant 77 feet per second squared. How many feet does the car travel before coming to a complete stop

Answer 1

144.6ft

Explanation:

Given

`u = 45ft/s` --- initial velocity

`a = -7ft/s^2` --- deceleration

Required

Distance traveled

To do this, we make use of:

`v^2 = u^2 + 2as`

Where

`v = 0` --- final velocity

So, we have:

`0^2 = 45^2 - 2*7*s`

`0 = 2025 - 14s`

Collect like terms

`14s = 2025`

Solve for s

`s = (2025)/(14)`

`s = 144.6`