The car's speed when it had traveled 17.0 m is 10.9 m/s
Explanation:
The motion of the car in this problem is a uniformly accelerated motion (=constant acceleration), therefore we can use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where:
u is the initial velocity of the car when it starts braking
v = 0 is the final velocity of the car (it comes to rest)
[tex]a=-3.50 m/s^2[/tex] is the deceleration of the car
s = 34.0 m is the distance covered by the car to stop
Solving for u, we find the initial velocity of the car:
[tex]u=\sqrt{-2as}=\sqrt{-2(-3.50)(34.0)}=15.4 m/s[/tex]
Now we can find the car's speed when it had traveled 17.0 m, by using again the same equation:
[tex]v^2-u^2=2as[/tex]
where this time we have:
v is the velocity we have to find
u = 15.4 m/s
[tex]a=-3.50 m/s^2[/tex]
s = 17.0 m
Solving for v,
[tex]v=\sqrt{u^2+2as}=\sqrt{(15.4)^2+2(-3.50)(17.0)}=10.9 m/s[/tex]
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