Question

A red toy car has an acceleration of 4.00 m/s², whereas a blue toy car has an acceleration of 2.00 m/s². You start the blue car from rest down a track of length 9.00 m. How much time must you wait before starting the red car from rest, such that the two cars reach the end of the track at the same time?

Answer 1

One must wait 0.88 secs before starting the red car

Step-by-step explanation:

To calculate how much time to wait before starting the red car from rest, we will calculate the time it will take the blue car to reach the end of the track from rest and the time it will take the red car to reach the end of the track from rest; then, the difference between these times is the time one must wait before starting the red car from rest such that two cars reach the end of the track at the same time

For the red car,

Acceleration = 4.00 m/s²

Initial velocity = 0 m/s (Since it is starting from rest)

Distance = 9.00 m

Let the time spent by the red car be
`t_(r)`

From one of the equations of kinematics for linear motion

`S = ut + (1)/(2)at^(2)`

Where

`S` is the distance traveled

`u` is the initial velocity

`t` is the time

and
`a` is the acceleration

Then, for the red car

`9.00 = (0)(t_(r)) + (1)/(2)(4.00)(t_(r))^(2)`

`9.00 = (2.00)(t_(r))^(2)`

`t_(r) = \sqrt{(9.00)/(2.00) }`

`t_(r) =2.12 secs`

This is the time it will take the red car to reach the end of the track

For the blue car

Acceleration = 2.00 m/s²

Initial velocity = 0 m/s (Since it is starting from rest)

Distance = 9.00 m

Let the time spent by the red car be
`t_(b)`

Also from

`S = ut + (1)/(2)at^(2)`

`9.00 = (0)(t_(r)) + (1)/(2)(2.00)(t_(b))^(2)`

`9.00 = (t_(b))^(2)`

`t_(b) = √(9)`

`t_(b) = 3secs`

This is the time it will take the blue car to reach the end of the track

The difference of the times is

`t_(b) - t_(r)` = 3 secs - 2.12 secs

= 0.88 secs

Hence, one must wait 0.88 secs before starting the red car