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]


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]

