A hockey puck sliding on a frozen lake comes to rest after traveling 160 m.
If its initial velocity is 2.3 m/s, what is its acceleration (assumed constant)?
Answer in units of m/s2.
To find the acceleration of the hockey puck, we can use the equation of motion:
vf^2 = vi^2 + 2aΔx
where:
- vf is the final velocity (0 m/s, as the puck comes to rest)
- vi is the initial velocity (2.3 m/s)
- a is the acceleration (unknown - what we're trying to find)
- Δx is the displacement (160 m)
Rearranging the equation, we have:
a = (vf^2 - vi^2) / (2Δx)
Substituting the given values:
a = (0^2 - 2.3^2) / (2 * 160)
Simplifying:
a = (-5.29) / 320
Dividing:
a ≈ -0.0165 m/s^2
Therefore, the acceleration of the hockey puck (assumed constant) is approximately -0.0165 m/s^2.
To find the acceleration of the hockey puck, we can use the equation of motion:
\(v^2 = u^2 + 2as\)
Where:
- \(v\) is the final velocity (0 m/s, since the puck comes to rest),
- \(u\) is the initial velocity (2.3 m/s),
- \(a\) is the acceleration, and
- \(s\) is the distance traveled (160 m).
Rearranging the equation, we have:
\(a = \frac{{v^2 - u^2}}{{2s}}\)
Substituting the given values, we get:
\(a = \frac{{(0 \, \text{m/s})^2 - (2.3 \, \text{m/s})^2}}{{2 \times 160 \, \text{m}}}\)
Simplifying the equation:
\(a = \frac{{0 - 5.29}}{{320}} \, \text{m/s}^2\)
Calculating the division:
\(a = -0.016 \, \text{m/s}^2\)
Therefore, the acceleration of the hockey puck is approximately -0.016 m/s².
The distance = (1/2)*(acceleration)*(time)^2 + (initial velocity)*(time)
substitute:
acceleration= (final velocity-initial velocity)/time
The distance = (1/2)*[(final velocity-initial velocity)/time]*(time)^2 + (initial velocity)*(time)
Find time & plug back into acceleration formula!