A rock is thrown straight up into the air with an initial speed of 53 m/s at time
t = 0.
Ignore air resistance in this problem. At what times does it move with a speed of 40 m/s? Note: There are two answers to this problem.
earlier time:
later time:
To determine the times at which the rock moves with a speed of 40 m/s, we can use the equations of motion.
Given:
Initial speed (u) = 53 m/s
Target speed (v) = 40 m/s
Acceleration (a) = -9.8 m/s^2 (assuming the direction of motion is upward, the acceleration due to gravity would be negative)
We can use the equation of motion to relate the final velocity (v) with the initial velocity (u), acceleration (a), and time taken (t):
v = u + at
Rearranging the equation to solve for time:
t = (v - u) / a
Let's plug in the values:
t = (40 - 53) / -9.8
t = -13 / -9.8
t ≈ 1.33 seconds
So, the rock reaches a speed of 40 m/s approximately 1.33 seconds after being thrown upwards.
Since there are two answers to this problem, we need to consider both the earlier and later times.
Earlier time:
To find the earlier time, we need to consider the rock's motion before reaching the speed of 40 m/s. At this point, the rock is still moving upwards, decelerating due to gravity. The speed 40 m/s must occur before the rock reaches its peak and starts to descend. Therefore, the earlier time occurs before the time we calculated earlier, at a point when the rock is still gaining height.
Later time:
The later time occurs after the rock has reached its peak and is starting to descend. The speed 40 m/s is achieved when the rock is on its way down.
So, to determine the exact times at which the rock moves with a speed of 40 m/s, we would need additional information such as the height at which the rock was thrown or the time it takes for the rock to reach its maximum height.