66.7 m/s2
66.7 m/s2
The change in velocity is 54 m/s - 5 m/s = 49 m/s. The time is 0.75 seconds. Using the equation a = (vf - vi) / t, the acceleration is (49 m/s) / (0.75 s) = 65.33 m/s^2.
A skydiver is increasing their speed during the first three seconds of free fall due to gravity pulling them downwards. As the skydiver falls, their speed will continue to increase until they reach terminal velocity.
Acceleration due to gravity is 9.8m/s/s, which is the same as 9.8m/s2. An acceleration of 9.8m/s/s means that with each passing second, the velocity of the skydiver increases by 9.8m/s. Therefore, after two seconds. a skydiver's velocity would be 19.6m/s. The acceleration will continue at 9.8m/s/s until the skydiver reaches terminal velocity, at which point the weight of the skydiver and the air resistance will be balanced, so the net force acting on the skydiver will be zero, at which point there will be no further acceleration.
The order of magnitude of the number of seconds in one hour is around 10^4, or 10,000 seconds.
the magnitude of the skydivers acceleration is zero as he is decelerating by opening his parachute!
66.7 m/s2
80 m/s 2 up
Maximum speed is about 220 to 230mph and can be achieved after about 20 seconds of freefall. Normal parachute opening speed should be not greater that 120mph to avoid damage to the parachute
(4 m/s - 54 m/s)/0.75 s = -50/0.75 m/s² = -200/3 ≈ - 66.67 m/s² (negative because he is decelerating)
If the 0.75 refers to seconds, then his acceleration is -66.66... (repeating) metres per second^2.
80 m/s2 up
66.7 m/s2
zero
The change in velocity is 54 m/s - 5 m/s = 49 m/s. The time is 0.75 seconds. Using the equation a = (vf - vi) / t, the acceleration is (49 m/s) / (0.75 s) = 65.33 m/s^2.
To calculate this, you divide the change in velocity, by the time.
If d = 16*t^2 then there is no significant air resistance.