Question

Question #b8d54

Answer 1

Here's what I got.

Explanation:

You can break down the motion of the car into two parts

• uniform acceleration with `a_1 = "0.5 m s"^(-2)` for a total of `"30 s"`
• uniform deceleration with `a_2 = "2 m s"^(-2)` until it comes to a complete stop

Let's start with the first part. You know that the car has an initial velocity of

`v_0 = "5 m s"^(-1)`

Now, an acceleration equal to `"0.5 m s"^(-2)` tells you that the velocity of the car increases, hence the positive sign, by `"0.5 m s"^(-1)` with every passing second.

This means that after `"30 s"`, the velocity of the car changed by

`30 color(red)(cancel(color(black)("s"))) * overbrace("0.5 m s"^(-1)/(1color(red)(cancel(color(black)("s")))))^(color(blue)("equaivalent to a"_1 = "0.5 m s"^(-2))) = "15 m s"^(-1)`

Therefore, you can say that the total velocity of the car at the end of the `"30 s"` is equal to

`v_"f 1" = "5 m s"^(-1) + "15 m s"^(-1) = "20 m s"^(-1)`

To calculate the distance traveled by the car in the first part of its motion, use the equation

`color(blue)(ul(color(black)(d_1 = v_0 * t_1 + 1/2 * a_1 * t_1^2)))`

Plug in your values to calculate `d_1`

`d_1 = "5 m" color(red)(cancel(color(black)("s"^(-1)))) * 30color(red)(cancel(color(black)("s"))) + 1/2 * "0.5 m" color(red)(cancel(color(black)("s"^(-2)))) * 30^2 color(red)(cancel(color(black)("s"^2)))`

`d_1 = "150 m" + "225 m" = "375 m"`

Now, let's move on to the second part of the motion. The first thing to mention here is that the deceleration of the car is equal to

`a_2 = "2 m s"^(-2)`

which implies a negative acceleration, i.e. an acceleration that opposes the current direction of movement. So you must use

`a_2 = - "2 m s"^(-2)`

in your calculations. Keep in mind that the minus sign is used here because we take the current direction of movement to be the positive.

So if we assume that the car is moving from `0` to the right of the `x` axis, then a negative acceleration acts to the left of the `x` axis.

So, a negative acceleration equal to `-"2 m s"^(-2)` tells you that the velocity of the car decreases, hence the negative sign, by `"2 m s"^(-1)` with every passing second.

This means that if the car starts with

`v_"f 1" = "20 m s"^(-1)`

which is the final velocity for the first part of the motion and the initial velocity for the second part of the motion, you will have

`20 color(red)(cancel(color(black)("m s"^(-1)))) * overbrace("1 s"/(2color(red)(cancel(color(black)("m s"^(-1))))))^(color(blue)("equivalent to |a"_2| = "2 m s"^(-2))) = "10 s"`

Therefore, it will take `"10 s"` for the car to come to a complete stop, i.e. for its final velocity to be equal to `"0 m s"^(-1)`.

The distance covered by the car in the second part of its motion will be

`d_2 = v_"f 1" * t_2 + 1/2 * a_2 * t_2^2`

Plug in your values to find -- do not forget that `a_2` carries a negative sign!

`d_2 = "20 m" color(red)(cancel(color(black)("s"^(-1)))) * 10color(red)(cancel(color(black)("s"))) + 1/2 * (-"2 m" color(red)(cancel(color(black)("s"^(-2))))) * 10^2 color(red)(cancel(color(black)("s"^(2))))`

`d_2 = "200 m" - "100 m" = "100 m"`

Therefore, you can say that the total distance covered by the car is equal to

`d_"total" = d_1 + d_2`

`d_"total" = "375 m" + "100 m" = color(darkgreen)(ul(color(black)("475 m")))`

and that the car traveled for a total time of

`t_"total" = t_1 + t_2`

`t_"total" = "30 s" + "10 s" = color(darkgreen)(ul(color(black)("40 s")))`