A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 81 m ahead of the vehicle. After a reaction time of t s, the ranger applies the brakes to produce an acceleration of −3.3 m/s2. What is the maximum reaction time al- lowed if the ranger is to avoid hitting the deer?

A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 81 m ahead of the vehicle. After a reaction time of t s, the ranger applies the brakes to produce an acceleration of −3.3 m/s2. What is the maximum reaction time al-
lowed if the ranger is to avoid hitting the deer? Answer in units of s.
Question 2, chap 2, sect 7.
part 1 of 2 10 points
A car starts from rest and travels for 6.8 s with a uniform acceleration of +2.0 m/s2. The driver then applies the brakes, causing a uniform acceleration of −1.2 m/s2. a) If the brakes are applied for 2.8 s, how
fast is the car going at the end of the braking period? Answer in units of m/s.
Question 3, chap 2, sect 7.
part 2 of 2 10 points
b) How far has it gone from its start? Answer in units of m.
Question 4, chap 2, sect 9.
part 1 of 2 10 points
A ball is dropped from rest at point O (height unknown). After falling for some time, it passes by a window of height 2.3 m and it does so during time tAB = 0.22 s. The acceleration of gravity is 9.8 m/s2 .
O
A
B
2.3 m
b b b b
b
b
b
b
b
b
b
b
x
y
The ball accelerates all the way down; let vA be its speed as it passes the window’s top A and vB its speed as it passes the window’s bottom B. How much did the ball speed up as it passed
the window; i.e., calculate ∆vdown = vB−vA ? Answer in units of m/s.
Question 5, chap 2, sect 9.
part 2 of 2 10 points
Calculate the speed vA at which the ball passes the window’s top. Answer in units of m/s.
Question 6, chap 2, sect 4.
part 1 of 5 10 points
The position versus time for a certain object moving along the x-axis is shown. The objects initial position is −3 m.
−6
−4
−2
2
4
6
8
10
0 1 2 3 4 5 6 7 8 9
b
b
b b
b
b
time (s)
p o si ti o n (m
)
Find the average velocity in the time inter- val 0 s to 1 s. Answer in units of m/s.
Question 7, chap 2, sect 4.
part 2 of 5 10 points
Find the average velocity in the time inter- val 0 s to 3 s. Answer in units of m/s.
Question 8, chap 2, sect 4.
part 3 of 5 10 points
homework 02 – KHALEEFOH, FAHAD – Due: Feb 9 2017, 11:00 pm (Central time) 2
Find the average velocity in the time inter- val 1 s to 3 s. Answer in units of m/s.
Question 9, chap 2, sect 4.
part 4 of 5 10 points
Find the average velocity in the time inter- val 3 s to 7 s. Answer in units of m/s.
Question 10, chap 2, sect 4.
part 5 of 5 10 points
Find the average velocity in the time inter- val 0 s to 9 s. Answer in units of m/s.
Question 11, chap 2, sect 6.
part 1 of 1 10 points
A tennis ball with a speed of 24.3 m/s is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 17.9 m/s. If the ball is in contact with the wall for
0.014 s, what is the average acceleration of the ball while it is in contact with the wall? Answer in units of m/s2.
Question 12, chap 2, sect 9.
part 1 of 2 10 points
A ball is thrown upward. Its initial ver- tical speed is 10.2 m/s and the acceleration of gravity is 9.8 m/s2 , as shown in the figure below. Neglect: Air resistance.
b
b
b
b
b
b
b b b b b b b
b
b
b
b
b
b
b
1 0 .2
m / s
9 .8
m / s2h m a x
What is its maximum height, hmax ? Answer in units of m.
Question 13, chap 2, sect 9.
part 2 of 2 10 points
Find the speed v A of the ball as the ball
passes a point A, which is at one quarter of
the maximum height hmax 4
.
Answer in units of m/s.
Question 14, chap 2, sect 4.
part 1 of 2 10 points
A person travels by car from one city to another. She drives for 29.5 min at 62.7 km/h, 11.4 min at 105 km/h, 36.8 min at 47.9 km/h, and spends 12.1 min along the way eating lunch and buying gas. Determine the distance between the cities
along this route. Answer in units of km.
Question 15, chap 2, sect 4.
part 2 of 2 10 points
Determine the average speed for the trip. Answer in units of km/h.

 

 

ANSWER

 t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

t is the reaction time

 t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.

PART 1

Part a)

Part b)

Explanation:

Since ball is dropped under uniform gravity

so here we can say that the distance of 3 m moved by the ball under uniform acceleration is given as

so we have

also we know that

now we will have

Part a)

Part b)

speed at the top of the window is

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