Total displacement of the car: 405 m
Explanation:
The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation
[tex]s_1=ut_1+\frac{1}{2}a_1 t_1^2[/tex]
where:
u = 0 is the initial velocity (the car starts from rest)
[tex]t_1 = 20 s[/tex] is the time elapsed in the 1st part
[tex]a_1=1.62 m/s^2[/tex] is the acceleration of the car in the 1st part
[tex]s_1[/tex] is the displacement of the car in the 1st part
Solving for [tex]s_1[/tex],
[tex]s_1=0+\frac{1}{2}(1.62)(20)^2=324 m[/tex]
We can also find the velocity of the car after these 20 seconds using the equation:
[tex]v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s[/tex]
Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:
[tex]s_2 = (\frac{v_1 + v_2}{2})t_2[/tex]
where:
[tex]v_1=32.4 m/s[/tex] is the initial velocity at the beginning of the 2nd phase
[tex]v_2=0[/tex] is the final velocity (the car comes to a stop)
[tex]t_2=5 s[/tex] is the time elapsed in the 2nd phase
Substituting,
[tex]s_2=\frac{32.4+0}{2}(5)=81 m[/tex]
So, the total displacement of the car is
[tex]s=s_1+s_2=324+81=405 m[/tex]
Learn more about accelerated motion:
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