1. By what factor does the stopping distance and stopping time change if you dou

Question

1. By what factor does the stopping distance and stopping time change if you double your driving speed? Assume your deceleration is constant. 2. You throw a ball straight up in the air. It goes up, and eventually it comes back down. You are fascinated by this amazing experiment, and you ponder the implications on the velocity and acceleration of the ba. Which of the following is a correct interpretation? (a) As the ball rises, its velocity and acceleration point in the same direction (b) As the ball rises, its velocity decreases and its acceleration decreases. (c) As the ball rises, its velocity decreases but its acceleration increases. (d) At the topmost point, the velocity of the ball is zero and its acceleration is zero. (e) As the ball rises, its velocity decreases, but its acceleration remains constant 3. A baseball player hits a homerun. At the highest point (a) the magnitude of the acceleration is zero. (b) the magnitude of the velocity is zero. (c) the magnitude of the velocity is the slowest. (d) more than one of the above is true. (e) none of the above is true. For the following problems, show all work to receive credit. The solution must be nea, clear, and self explained. All numerical ansuwers must have appropriate number of significant figures. All numerical answers must have units, if appropriate. Answers without calculations will receive zero for the problem . A fireman 50.0 m away from a burning building directs a stream of water from ground-level fire hose at an angle of 30.0° above the horizontal. If the speed of the stream as it leaves the hose is vi = 40.0 m%, at what height will the stream of water strike the building?

Explanation / Answer

1.

Suppose Initial Velocity is V0 and de-acceleration is '-a', then stopping distance will be given by:

final velocity will be zero

V^2 = U^2 + 2*a*d

d = (0^2 - V0^2)/(2*(-a))

d = V0^2/(2a)

Stopping time will be

V = U + a*t

0 = V0 - a*t

t = V0/a

Now when initial speed is doubled, then Stopping distance will be

V^2 = U^2 + 2*a*d1

d1 = (0^2 - (2*V0)^2)/(2*(-a))

d1 = 2*V0^2/a = 4*(V0^2/2a) = 4*d

So Stoping distance will be quadrupled. (4 times)

Now new stopping time will be

V = U + a*t1

0 = 2*V0 - a*t1

t1 = 2*V0/a = 2*(V0/a) = 2*t

So stopping time will be doubled. (2 times)

2.

When ball is thrown upward then

At the lowest point when ball is thrown, velocity will be highest and upward

After that velocity will start decreasing and at max height velocity will be zero.

After that ball will stat traveling downward and velocity will start to increase.

About the acceleration: Acceleration remains constant through whole motion and will be equal to g (9.81 m/sec^2) and direction will always be downward.

So from this we can see that Correct option is E.

3.

At the highest point in projectile motion

Again like question 2, at the highest point vertical velocity will be zero, but it will have horizontal speed.

In projectile motion horizontal velocity remains constant because there is no acceleration in horizontal

direction. So since at every other there will be both vertical and horizontal velocity in projectile So velocity

will be lowest at highest because at that time there will be only horizontal speed.

Again Vertical acceleraion will be equal to g(9.81 m/sec^2) and downward.

So from this we can see that Correct option is C.

Please Upvote.

Comment below if you have any doubt.

See that I've answered 3 different question(1 as per chegg policy). So please ask the 4th question as a new question. I will be happy to help.

Please Upvote.

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