When you take your 1900-kg car out for a spin, you go around a corner of radius 53m with a speed of 13m/s. The coefficient of st
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Answer:
Approximately , assuming friction between the vehicle and the ground is negligible.
Explanation:
Let denote the mass of the vehicle. Let
denote the initial velocity of the vehicle. Let
denote the spring constant (needs to be found.) Let
denote the maximum displacement of the spring.
Convert velocity of the vehicle to standard units (meters per second):
.
Initial kinetic energy () of the vehicle:
.
When the vehicle is brought to a rest, the elastic potential energy () stored in the spring would be:
.
By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial of the vehicle should be equal to the
of the vehicle. In other words:
.
Rearrange this equation to find an expression for , the spring constant:
.
Substitute in the given values ,
, and
:
Answer:
If the acceleration is constant, the movements equations are:
a(t) = A.
for the velocity we can integrate over time:
v(t) = A*t + v0
where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:
Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.