Question

A truck traveling at a constant speed of 40.0 km/h applies its brakes and comes to a complete stop in 5.0 s.

a. Convert the truck's constant speed from km/h to m/s.
b. Draw a simple v-t graph of the truck stopping. Use v = m/s.
c. Calculate the average acceleration for the truck. What does this value mean?
d. The truck starts again and accelerates at a constant rate of 0.80 m/s2. Beginning at the stopping point of the truck, extend the graph to show 10.0 s of this new constant acceleration.
e. After restarting, how much time does it take for the truck to regain its original speed of 40.0 km/h?
f. After restarting, how far does the truck travel before reaching its original speed of 40.0 km/h?

Answer 1

a)
`v = 11.111\,(m)/(s)`, b) See attachment below, c)
`a = - 2.222\,(m)/(s^(2))`. It means that speed is decreasing at an average rate of 2.222 m/s², d) See attachment below, e)
`\Delta t = 13.889\,s`, f)
`\Delta s = 77.159\,m`.

Step-by-step explanation:

a) The constant speed of the truck is:

`v = \left(40\,(km)/(h) \right)\cdot \left((1\,h)/(3600\,s) \right)\cdot \left((1000\,m)/(1\,km) \right)`

`v = 11.111\,(m)/(s)`

b) A simple graph is presented as attachment.

c) The average acceleration for the truck is:

`a = (0\,(m)/(s) - 11.111\,(m)/(s) )/(5\,s - 0\,s)`

`a = - 2.222\,(m)/(s^(2))`

It means that speed is decreasing at an average rate of 2.222 m/s².

d) The new graph is present as attachment.

e) The time needed to regain the original speed is:

`\Delta t = (11.111\,(m)/(s) )/(0.8\,(m)/(s^(2)) )`

`\Delta t = 13.889\,s`

f) The distance travelled by the truck is:

`\Delta s = ((11.111\,(m)/(s) )^(2))/(2\cdot (0.80\,(m)/(s^(2)) ))`

`\Delta s = 77.159\,m`

Answer 2

Part a)

`v = 11.11 m/s`

Part c)

`a = -2.22 m/s^2`

This mean the truck is decelerating and its speed is decreasing

Part e)

`t = 13.89 s`

Part f)

`d = 77.14 m`

Step-by-step explanation:

Part a)

Speed of the truck is given as

`v = 40 km/h`

as we know that

1 km = 1000 m

1 h = 3600 s

so we will have

`v =40 * (1000)/(3600) m/s`

`v = 11.11 m/s`

Part c)

Average acceleration is given as

`a = (v_f - v_i)/(t)`

now we have

`a = (0 - 11.11)/(5)`

`a = -2.22 m/s^2`

Part e)

as we know that it is uniform acceleration

so we can say

`v_f - v_i = at`

`11.11 - 0 = 0.80 t`

`t = 13.89 s`

Part f)

distance traveled by the truck is given as

`d = (1)/(2)at^2`

`d = (1)/(2)(0.80)(13.89^2)`

`d = 77.14 m`