Answer:
a. 
b.
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
is the time taken to stop after braking
Explanation:
Given:
- speed of leading car,

- speed of lagging car,

- distance between the cars,

- deceleration of the leading car after braking,

a.
Time taken by the car to stop:

where:
, final velocity after braking
time taken


b.
using the eq. of motion for the given condition:

where:
final velocity of the chasing car after braking = 0
acceleration of the chasing car after braking

must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
time taken by the chasing car to stop:


is the time taken to stop after braking
The value of x in the given right triangle in a semicircle is determined as 21.
<h3>What is the measure of a triangle in a semicircle?</h3>
The triangle in a semicircle is always a right angle triangle.
From the figure shown, we can say that the triangle G J K is right triangle and m<K = 90degrees.
Given that m<K = 4x + 6, we will can use the following equation to find the value of x as shown:
4x + 6 = 90
4x = 90 - 6
4x = 84
x = 21
Thus, the value of x in the given right triangle in a semicircle is determined as 21.
Learn more about right angle here: brainly.com/question/64787
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Answer:
i think the answer is constant
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>