A diver enters the water at a speed of 28.2 m/s and stops after traveling 4.00 m. Assuming uniform acceleration, we need to calculate: a) the average acceleration of the diver. b) the time it takes for the diver to stop. c) the speed of the diver after traveling 2.00 m.

1. Problem Description

A diver enters the water at a speed of 28.2 m/s and stops after traveling 4.00 m. Assuming uniform acceleration, we need to calculate:

a) the average acceleration of the diver.

b) the time it takes for the diver to stop.

c) the speed of the diver after traveling 2.00 m.

2. Solution Steps

a) Calculate the average acceleration.

We can use the following kinematic equation:

$v_f^2 = v_i^2 + 2 a \Delta x$

where $v_f$ is the final velocity, $v_i$ is the initial velocity, $a$ is the acceleration, and $\Delta x$ is the displacement.

In this case, $v_f = 0$ m/s, $v_i = 28.2$ m/s, and $\Delta x = 4.00$ m.

Solving for $a$:

$0^2 = (28.2)^2 + 2 a (4.00)$

$0 = 795.24 + 8 a$

$a = -795.24 / 8$

$a = -99.405$ m/s$^2$

b) Calculate the time it takes to stop.

We can use the following kinematic equation:

$v_f = v_i + a t$

where $t$ is the time.

In this case, $v_f = 0$ m/s, $v_i = 28.2$ m/s, and $a = -99.405$ m/s$^2$.

Solving for $t$:

$0 = 28.2 + (-99.405) t$

$99.405 t = 28.2$

$t = 28.2 / 99.405$

$t = 0.2837$ s

c) Calculate the speed after diving 2.00 m.

Using the same kinematic equation as in part a:

$v_f^2 = v_i^2 + 2 a \Delta x$

In this case, $v_i = 28.2$ m/s, $a = -99.405$ m/s$^2$, and $\Delta x = 2.00$ m.

Solving for $v_f$:

$v_f^2 = (28.2)^2 + 2 (-99.405) (2.00)$

$v_f^2 = 795.24 - 397.62$

$v_f^2 = 397.62$

$v_f = \sqrt{397.62}$

$v_f = 19.94$ m/s

3. Final Answer

a) The average acceleration of the diver is -99.405 m/s$^2$.

b) The time it takes for the diver to come to a stop is 0.2837 s.

c) The speed of the diver after diving for 2.00 m is 19.94 m/s.