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β 11y ago66.7 m/s2
Alex Kuhic
Using the equation for acceleration (a = \frac{vf-vi}{t}), we find the deceleration to be (a = \frac{4-54}{0.75}) which equals -70 m/s(^2). Since the acceleration is negative, it indicates deceleration.
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.
A skydiver's velocity after 2 seconds will depend on factors such as their initial velocity, weight, air resistance, and gravitational force acting on them. On average, a skydiver will reach a terminal velocity of around 120 mph (193 km/h) after about 10 seconds of freefall.
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.
66.7 m/s2
80 m/s2 up
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.