Question

A car originally traveling at 30.0 m/s manages to brake for 5.0 seconds while traveling 125 m along a road. After those first 5.0 seconds, the brakes fail. After an additional 5.0 seconds it travels an additional 150 m further down the road. What was the magnitude of the acceleration of the car after the brakes failed

Answer 1

The magnitude of the acceleration of the car after the brakes failed is 4 m/s²

Step-by-step explanation:

The car was originally traveling at 30.0 m/s, that is

The initial velocity,
`u` = 30.0 m/s

The time spent while the car manages to brake is 5.0 seconds, that is

time,
`t` = 5.0 secs

and the distance traveled during this time is

distance,
`s` = 125 m

From one of the equations of kinematics for linear motion,

`s = ut + (1)/(2)at^(2) \\`

Where
`a` is the acceleration

We can determine the deceleration of the car during the first 5.0 seconds

Hence,

From,

`s = ut + (1)/(2)at^(2) \\`

`125 = 30.0(5.0) + (1)/(2)(a)(5.0)^(2)`

`125 =150.0 + 12.5a`

`12.5a = 125 - 150.0`

`12.5a = -25\\a = (-25)/(12.5)\`

`a = - 2.0 m/s^(2)` (Negative sign indicates deceleration)

Now we will calculate the final velocity reached at this time

From,

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

Where
`v` is the final velocity

`v^(2) = 30.0^(2) + 2(-2.0)(125)\\v^(2) = 400\\v = √(400) \\v = 20 m/s \\`

This is the final velocity reached by the car during the first 5.0 seconds

Now, for the magnitude of the acceleration of the car after the brakes failed,

After the brakes failed,

it travels an additional 150 m further down the road, that is

s = 150m

an additional 5.0 seconds, that is

t = 5.0 seconds

Also, from

`s = ut + (1)/(2)at^(2) \\`

The initial velocity here will be the final velocity for the first 5.0 seconds, that is,

u = 20 m/s

Hence,

`s = ut + (1)/(2)at^(2) \\` becomes

`150 = 20(5.0) + (1)/(2)(a)(5.0)^(2)`

`150 = 100 + 12.5a\\12.5a = 150 - 100\\12.5a = 50\\a = (50)/(12.5) \\a = 4m/s^(2)`

Hence, the magnitude of the acceleration of the car after the brakes failed is 4 m/s²