An 865 kg (1912 lb) dragster, starting from rest, attains a speed of 26.7 m/s (59.8 mph) in 0.59 s.

(a) Find the average acceleration of the dragster during this time interval.
m/s2
(b) What is the magnitude of the average net force on the dragster during this time?
N
(c) Assume that the driver has a mass of 68 kg. What horizontal force does the seat exert on the driver?
N

acceleration= changevelocty/time

Forceaverage= mass*acceleraltion

Forceseat= driver mass*acceleration

A 69 kg diver jumps off a 10.4 m tower.

(a) Find the diver's velocity when he hits the water.
m/s
(b) The diver comes to a stop 2.0 m below the surface. Find the net force exerted by the water.
N

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To solve this problem, we will use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

(a) To find the average acceleration of the dragster, we can use the equation:

acceleration = (final velocity - initial velocity) / time

Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = 26.7 m/s
Time (t) = 0.59 s

Substituting the values into the equation, we get:

acceleration = (26.7 m/s - 0 m/s) / 0.59 s
acceleration = 45.254 m/s^2

Therefore, the average acceleration of the dragster during this time interval is 45.254 m/s^2.

(b) To find the magnitude of the average net force on the dragster, we can use Newton's second law of motion. Rearranging the equation, we have:

force = mass * acceleration

Given:
Mass (m) = 865 kg
Acceleration (a) = 45.254 m/s^2

Substituting the values into the equation, we get:

force = 865 kg * 45.254 m/s^2
force = 39,152.91 N

Therefore, the magnitude of the average net force on the dragster during this time is approximately 39,152.91 N.

(c) To find the horizontal force exerted by the seat on the driver, we need to consider the gravitational force acting on the driver and the net force acting on the driver.

The gravitational force is given by the equation:

gravitational force = mass of the driver * gravitational acceleration

Given:
Mass of the driver (md) = 68 kg
Gravitational acceleration (g) = 9.8 m/s^2

Substituting the values into the equation, we get:

gravitational force = 68 kg * 9.8 m/s^2
gravitational force = 666.4 N

Since the driver is not accelerating vertically, the net force in the vertical direction is zero. Therefore, the vertical force exerted by the seat on the driver is equal to the gravitational force.

In the horizontal direction, there is no acceleration, so the net horizontal force is also zero. Therefore, the horizontal force exerted by the seat on the driver is also zero.

Therefore, the horizontal force exerted by the seat on the driver is zero N.