1) The car overtakes the truck at a distance of 160 m far from the intersection.
2) The velocity of the car is 40 m/s
Explanation:
1)
The car is travelling with a constant acceleration starting from rest, so its position at time t (measured taking the intersection as the origin) is given by
[tex]x_c(t) = \frac{1}{2}at^2[/tex]
where
[tex]a=5 m/s^2[/tex] is the acceleration
t is the time
On the other hand, the truck is travelling at a constant velocity, therefore its position at time t is given by
[tex]x_t(t) = vt[/tex]
where
v = 20 m/s is the velocity of the truck
t is the time
The car overtakes the truck when the two positions are the  same, so when
[tex]x_c(t) = x_t(t)\\\frac{1}{2}at^2 = vt\\t=\frac{2v}{a}=\frac{2(20)}{5}=8 s[/tex]
So, after a time of 8 seconds. Therefore, the distance covered by the car during this time is
[tex]x_c(8) = \frac{1}{2}(5)(8)^2=160 m[/tex]
So, the car overtakes the truck 160 m far from the intersection.
2)
The motion of the car is a uniformly accelerated motion, so the velocity of the car at time t is given by the suvat equation
[tex]v=u+at[/tex]
where
v is the velocity at time t
u is the initial velocity
a is the acceleration
For the car in this problem, we have:
u = 0 (it starts  from rest)
[tex]a=5 m/s^2[/tex]
And we know that the car overtakes the truck when
t = 8 s
Substituting into the equation,
[tex]v=0+(5)(8)=40 m/s[/tex]
Learn more about accelerated motion:
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