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?

Respuesta :

Answer:

Part a)

[tex]v = 11.11 m/s[/tex]

Part c)

[tex]a = -2.22 m/s^2[/tex]

This mean the truck is decelerating and its speed is decreasing

Part e)

[tex]t = 13.89 s[/tex]

Part f)

[tex]d = 77.14 m[/tex]

Explanation:

Part a)

Speed of the truck is given as

[tex]v = 40 km/h[/tex]

as we know that

1 km = 1000 m

1 h = 3600 s

so we will have

[tex]v =40 \times \frac{1000}{3600} m/s[/tex]

[tex]v = 11.11 m/s[/tex]

Part c)

Average acceleration is given as

[tex]a = \frac{v_f - v_i}{t}[/tex]

now we have

[tex]a = \frac{0 - 11.11}{5}[/tex]

[tex]a = -2.22 m/s^2[/tex]

Part e)

as we know that it is uniform acceleration

so we can say

[tex]v_f - v_i = at[/tex]

[tex]11.11 - 0 = 0.80 t[/tex]

[tex]t = 13.89 s[/tex]

Part f)

distance traveled by the truck is given as

[tex]d = \frac{1}{2}at^2[/tex]

[tex]d = \frac{1}{2}(0.80)(13.89^2)[/tex]

[tex]d = 77.14 m[/tex]

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Ver imagen aristocles

Answer:

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

Explanation:

a) The constant speed of the truck is:

[tex]v = \left(40\,\frac{km}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)\cdot \left(\frac{1000\,m}{1\,km} \right)[/tex]

[tex]v = 11.111\,\frac{m}{s}[/tex]

b) A simple graph is presented as attachment.

c) The average acceleration for the truck is:

[tex]a = \frac{0\,\frac{m}{s} - 11.111\,\frac{m}{s} }{5\,s - 0\,s}[/tex]

[tex]a = - 2.222\,\frac{m}{s^{2}}[/tex]

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:

[tex]\Delta t = \frac{11.111\,\frac{m}{s} }{0.8\,\frac{m}{s^{2}} }[/tex]

[tex]\Delta t = 13.889\,s[/tex]

f) The distance travelled by the truck is:

[tex]\Delta s = \frac{(11.111\,\frac{m}{s} )^{2}}{2\cdot (0.80\,\frac{m}{s^{2}} )}[/tex]

[tex]\Delta s = 77.159\,m[/tex]

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Ver imagen xero099
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