a rock hits the ground at a speed of 10m/s and leaves a hole 25cm deep. what is the magnitude of the (assumed) uniform deceleration of the rock?
Vf^2=Vi^2 + 2ad solve for a.
2,000 m/s^2
To find the magnitude of the uniform deceleration of the rock, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (0 m/s since the rock comes to rest)
- vi is the initial velocity (10 m/s)
- a is the uniform deceleration we are trying to find
- d is the displacement (25 cm = 0.25 m)
Let's plug in the given values:
0 = (10)^2 + 2a(0.25)
Simplifying the equation:
0 = 100 + 0.5a
Rearranging the equation to solve for "a":
0.5a = -100
a = -100 / 0.5
a = -200 m/s^2
Hence, the magnitude of the assumed uniform deceleration of the rock is 200 m/s^2.
To find the magnitude of the deceleration of the rock, we can use the equation of motion:
\(v^2 = u^2 + 2as\)
where:
- \(v\) is the final velocity (which is 0 as the rock comes to a stop),
- \(u\) is the initial velocity (10 m/s),
- \(a\) is the deceleration (what we are trying to find), and
- \(s\) is the displacement (25 cm or 0.25 m, since it's given in centimeters).
Let's rearrange the equation to solve for the deceleration:
\(0 = (10 \, \text{m/s})^2 + 2a \cdot (0.25 \, \text{m})\)
\(0 = 100 \, \text{m/s}^2 + 0.5a\)
Rearranging again:
\(0.5a = -100 \, \text{m/s}^2\)
\(a = \frac{-100 \, \text{m/s}^2}{0.5}\)
\(a = -200 \, \text{m/s}^2\)
Therefore, the magnitude of the uniform deceleration of the rock is 200 m/s². Note that the negative sign indicates that the deceleration is in the opposite direction of the initial velocity.