When a parachute opens, the air exerts a large drag force on it. this upward force is initially greater than the weight of the sky diver and, thus, slows him down. suppose the weight of the sky diver is 915 n and the drag force has a magnitude of 1027 n. the mass of the sky diver is 93.4 kg. what are the magnitude and direction of his acceleration?

The magnitude of the acceleration is 1.2 m/s2 and the direction is positive, upward.

Explanation:

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky driver, and, thus, slow him down. It means, if the drag force is greater than the weight, acceleration should be upward and positive to slows him.

We have the weight W = 915 N of the sky driver. This is negative, due to the direction.

The drag force D = 1027 N. This is positive, due to the direction upward. The mass of the sky driver is m=93.4 kg with an unknown acceleration a.

The Newton laws shoued us:

∑F = m*a

D - W = m*a

1027 - 915 = 93.4*a

93.4*a = 112

a = 112/93.4

a = 1.2 m/s2

Finally, we can conclude that The magnitud of the acceleration is 1.2 m/s2 and the direction is positive, upward.

seeing the graph would

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s