Question

Question #706f8

Answer 1

It accelerates well.

Explanation:

In simple terms, acceleration, `a`, tells you how the velocity of an object changes with respect to time.

`color(purple)(bar(ul(|color(white)(a/a)color(black)(a = (Deltav)/(Deltat)color(white)(a/a)|)))`

Here

`Deltav` - the change in the velocity of the object
`Deltat` - the time needed for this change to take place

The problem tells you that a car goes from `"0 mi/h"` to `"60 mi/h"`, so the change in the velocity of the car, `Deltav`, is equal to

`Deltav = "60 mi/h" - "0 mi/h" = "60 mi/h"`

You also know that it takes `6` seconds for this change in velocity to take place. If you take `t=0` as the time when the car begins to accelerate, you can say that you have

`Deltat = "6 s" - "0 s" = "6 s"`

Since you know the values of `Deltav` and `Deltat`, you can say that you know the car's acceleration.

To calculate it, you must convert the change in velocity to meters per second by using a series of conversion factors

`60 color(red)(cancel(color(black)("mi")))/(1color(red)(cancel(color(black)("h")))) * (1.61 color(red)(cancel(color(black)("km"))))/(1color(red)(cancel(color(black)("mi")))) * (10^3"m")/(1color(red)(cancel(color(black)("km")))) * (1color(red)(cancel(color(black)("h"))))/(60color(red)(cancel(color(black)("min")))) * (1color(red)(cancel(color(black)("min"))))/"60 s" = "26.83 m s"^(-1)`

You will have

`a = ("26.83 m s"^(-1))/"6 s" ~~ "4.5 m s"^(-2)`

What this tells you is that with every passing second, the velocity of the car increases by approximately `"4.5 m s"^(-1)`.

Therefore, because you know the change in velocity and the time needed for that change to take place, you can say that the car accelerates well.

It can be argued that you can also say that the car is fast, but for that to make a valid statement you'd need to know its top speed. In relative terms, a car that accelerates that well will most likely be very fast, but I don't think that this is the conclusion you must draw here.