A hockey puck sliding on a frozen lake comes to rest after traveling 265 m.
If its initial velocity is 4.2 m/s, what is its acceleration (assumed constant)?
Answer in units of m/s2.
A hockey puck sliding on a frozen lake comes
to rest after traveling 302 m.
If its initial velocity is 2.9 m/s, what is its
acceleration (assumed constant)?
Answer in units of m/s
2
.
To find the acceleration of the hockey puck, we need to use the equation of motion:
v^2 = u^2 + 2as
where:
- v is the final velocity (0 m/s since it comes to rest)
- u is the initial velocity (4.2 m/s)
- a is the acceleration (unknown)
- s is the distance traveled (265 m)
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0^2 - 4.2^2) / (2 * 265)
Simplifying:
a = (-17.64) / (530)
a ≈ -0.033 m/s^2
Therefore, the acceleration of the hockey puck is approximately -0.033 m/s^2.
To find the acceleration of the hockey puck, we can use the equation of motion,
v_f^2 = v_i^2 + 2*a*d
where:
v_f is the final velocity (which is 0 since the puck comes to rest),
v_i is the initial velocity,
a is the acceleration, and
d is the distance traveled.
Since v_f = 0, we can rearrange the equation to solve for acceleration (a):
0 = v_i^2 + 2*a*d
Substituting the given values:
v_i = 4.2 m/s
d = 265 m
0 = (4.2)^2 + 2*a*(265)
Simplifying the equation,
0 = 17.64 + 530a
To solve for a, let's isolate the term with a:
530a = -17.64
a = -17.64 / 530
a ≈ -0.0332 m/s^2
Therefore, the acceleration of the hockey puck (assumed as constant) is approximately -0.0332 m/s^2.
Vf^2 = Vo^2 + 2ad,
0^2 = (4.2)^2 + 2 * a * d,
0 = 17.64 + 530a,
530a = -17.64,
a = -17.64 / 530 = -0.0333 m/s^2.