The car hits the kangaroo at a velocity of 6.3 m/s.
Explanation:
In the first 0.6 seconds (reaction time), the car continues travelling at a constant velocity of
u = 30 m/s
Therefore, since the time interval is
t = 0.6 s
It means that the car covers a distance of
[tex]d_1 = ut=(30)(0.6)=18 m[/tex]
The initial distance from the kangaroo was
D = 65 m
Therefore, the distance left to the kangaroo is
[tex]d=D-d_1 = 65-18=47 m[/tex]
In this second part, the car travels with a uniformly accelerated motion, therefore we can use the suvat equation
[tex]v^2-u^2 = 2ad[/tex]
where:
v is the final velocity of the car when it reaches the kangaroo
u = 30 m/s is the initial velocity
[tex]a=-10 m/s^2[/tex] is the deceleration
s = 47 m is the distance to be cover
Solving for v, we find:
[tex]v=\sqrt{u^2+2ad}=\sqrt{30^2+2(-10)(47)}=6.3 m/s[/tex]
So, the velocity of the car when it hits the kangaroo is 6.3 m/s.
Learn more about accelerated motion:
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