# Question #b8d54

##### 1 Answer

Here's what I got.

#### Explanation:

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

uniform accelerationwith#a_1 = "0.5 m s"^(-2)# for a total of#"30 s"# uniform decelerationwith#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 **increases**, hence the *positive sign*, by **with every passing second**.

This means that after *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

#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 = "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 **right** of the **to the left** of the

So, a negative acceleration equal to **decreases**, hence the *negative sign*, by **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

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 *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")))#