Question

Question #76c41

Answer 1

`"90 m"`

Explanation:

Your goal here is to find the acceleration of the car so that you can use the equation

`color(blue)(ul(color(black)(d = v_0 * t + 1/2 * a * t^2)))`

Here

• `d` is the distance traveled in the `"3 s"`
• `v_0` is the initial velocity of the car
• `t` is the time given
• `a` is the acceleration of the car

Now, you know that an object's acceleration tells you how said object's velocity is changing with respect to time.

`a = (Deltav)/(Deltat) color(white)(color(blue)( larr " change in velocity")/(color(purple)(larr" change in time"))`

In your case, you know that the velocity of the car changes by

`Deltav = "25 m s"^(-1) - "15 m s"^(-1)`

`Deltav = "10 m s"^(-1)`

and that it takes `"3 s"` for this change to occur. If you take `t=0` to be the time when the car begins to accelerate, you can say that the change in time is equal to

`Deltat = "3 s"- "0 s"`

`Deltat = "3 s"`

This means that the car's acceleration is equal to

`a = "10 m s"^(-1)/"3 s" = 10/3 color(white)(.)"m s"^(-2)`

This means that with every passing second, the velocity of the car increases by `10/3color(white)(.)"m s"^(-1)`.

Plug this into the first equation and calculate `d`

`d = "15 m" color(red)(cancel(color(black)("s"^(-1)))) * 3color(red)(cancel(color(black)("s"))) + 1/2 * 10/3color(white)(.)"m" color(red)(cancel(color(black)("s"^(-2)))) * 3^2 color(red)(cancel(color(black)("s"^(2))))`

`d = "45 m" + "45 m" = color(darkgreen)(ul(color(black)("90 m")))`

Therefore, you can say that the car travels `"90 m"` in the `"3 s"` it's being accelerated.