To solve the given problem, we'll assume that the rock falls freely under the influence of gravity, neglecting air resistance. Let's calculate the rock's acceleration, distance traveled in 5 seconds, and its velocity after 10 seconds.
- Acceleration (a): We can determine the acceleration using the equation of motion for uniformly accelerated linear motion:
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given: Initial velocity (u) = 10 m/s Final velocity (v) = 40 m/s Time (t) = 2 seconds
Rearranging the equation, we have:
a = (v - u) / t a = (40 m/s - 10 m/s) / 2 s a = 30 m/s / 2 s a = 15 m/s²
So, the acceleration of the rock is 15 m/s².
- Distance traveled in 5 seconds: We can use the equation of motion for distance:
s = ut + (1/2)at²
where: s is the distance, u is the initial velocity, t is the time, and a is the acceleration.
Given: Initial velocity (u) = 10 m/s Time (t) = 5 seconds Acceleration (a) = 15 m/s²
Plugging in the values, we get:
s = (10 m/s) * 5 s + (1/2) * (15 m/s²) * (5 s)² s = 50 m + (1/2) * 15 m/s² * 25 s² s = 50 m + 187.5 m s = 237.5 m
Therefore, the rock travels a distance of 237.5 meters in 5 seconds.
- Velocity after 10 seconds: The final velocity after 10 seconds can be found using the equation of motion:
v = u + at
Given: Initial velocity (u) = 10 m/s Time (t) = 10 seconds Acceleration (a) = 15 m/s²
Plugging in the values, we get:
v = 10 m/s + (15 m/s²) * 10 s v = 10 m/s + 150 m/s v = 160 m/s
Hence, the rock's velocity after 10 seconds of falling is 160 m/s.