Motion at Constant Acceleration Mr. Kepple

Name: _______________________
Mr. Kepple
Motion at Constant Acceleration
1D Kinematics – HW#5
Date: ___________ Period: _____
1. A car accelerates from 12 m/s to 21 m/s in 6.0 s. (a) What was its acceleration? (b) How far did it
travel in this time? Assume constant acceleration.
(b) Distance traveled
(a) Acceleration
∆𝑣 2 − 2
𝑎=
=
=
∆𝑡
6
5 m/s
2
1
𝑥 − 𝑥0 = 𝑣0 𝑡 + 𝑎𝑡 2
2
𝑥−
1
= ( 2)(6) + ( 5)(6)2
2
𝑥 = 99 m
2. Suppose a rocket ship in deep space moves with constant acceleration equal to 9.8 m/s², which gives
the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a
speed one-tenth that of light, which travels at
m/s? (b) How far will it travel in doing so?
(a) Time until 𝑣 =
𝑐
𝑣 = 𝑣0 + 𝑎𝑡
1
𝑥 − 𝑥0 = 𝑣0 𝑡 + 𝑎𝑡 2
2
𝑣 − 𝑣0
𝑡=
𝑎
𝑡=
(
𝑥−
)(
)−
98
6
𝑡=
(b) Distance traveled
=
𝑥 = 46
1
+ (9 8)(
2
13
6 2
)
m
s
3. A world-class sprinter can reach a top speed (of about 11.5 m/s) in the first 15.0 m of the race. (a)
What is the average acceleration of this sprinter and (b) how long does it take her to reach that speed?
(a) Average Acceleration
(b) Time
𝑣 2 = 𝑣0 2 + 2𝑎(𝑥 − 𝑥0 )
𝑣 = 𝑣0 + 𝑎𝑡
𝑣 2 − 𝑣0 2
𝑎=
2(𝑥 − 𝑥0 )
𝑡=
5)2 −
2( 5)
𝑡=
𝑎=
(
= 4 4 m/s²
𝑣 − 𝑣0
𝑎
5−
= 26 s
44 8
4. In coming to a stop, a car leaves ski marks 85 m long on the highway. Assuming a deceleration of
4.00 m/s², estimate the speed of the car just before braking.
𝑣 2 = 𝑣0 2 + 2𝑎(𝑥 − 𝑥0 )
𝑣 2 − 2𝑎(𝑥 − 𝑥0 )
𝑣0 =
𝑣0 =
− 2(−4)(85) = 26 m/s
5. An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. His car is capable of
decelerating at a rate of 3.65 m/s². If it takes him 0.200 s to get the brakes on and he is 45.0 m from
the intersection when he sees the light, will he be able to stop in time? (4 points)
Distance during reaction time
Distance to stop
𝑥 = 𝑣𝑡
𝑥 = ( 8)( 2) =
𝑣 2 = 𝑣0 2 + 2𝑎(𝑥 − 𝑥0 )
6m
𝑣 2 − 𝑣0 2
∆𝑥 =
2𝑎
Total distance
𝑑=
− ( 8)2
∆𝑥 =
= 44 4 m
2(− 65)
6 + 44 4 = 48 m
The driver will NOT stop in time.
6. An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s² in a straight line until it
reaches a speed of 20 m/s. The vehicle then slows at a constant rate of 1.0 m/s² until it stops. (a) How
much time elapses from start to stop? (b) How far does the vehicle travel from start to stop? (6 points)
(a) Time elapsed
𝑣 = 𝑣0 + 𝑎𝑡
𝑡=
𝑣 − 𝑣0
𝑎
(a) Distance traveled
𝑣 2 = 𝑣0 2 + 2𝑎(𝑥 − 𝑥0 )
𝑣 2 − 𝑣0 2
∆𝑥 =
2𝑎
𝑡1 =
(2 ) −
(2)
=
s
𝑡 = 𝑡1 + 𝑡2 =
(2 )2 −
𝑥1 =
2(2)
𝑥2 =
=
− (2 )2
=2
2(− )
𝑥 = 𝑥1 + 𝑥2 =
− (2 )
=2 s
(− )
𝑡2 =
+2 =
s
m
m
+2
=
m
Name: _______________________
Mr. Kepple
Freely Falling Objects
1D Kinematics – HW#6
Date: ___________ Period: _____
1. A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.75 s. How high is
the cliff?
Take downward to be
the positive direction.
𝑦
Choose 𝑦
at the
top of the cliff.
𝑦
2. If a car rolls gently (
𝑣 + 𝑎𝑡
𝑡
𝑣−𝑣
𝑎
2
+ +
𝑦
1
2
9.8 3.75
2
68.9 m
off a vertical cliff, how long does it take to reach 55 km/h? (3 points)
Take downward to be
the positive direction.
𝑣
1
𝑦 + 𝑣 𝑡 + 𝑎𝑡 2
𝑣
𝑡
55 km 1 3 m
1h
×
×
1h
1 km 36 s
15.28 −
9.8
15.28 m/s
1.6 s
3. A ball player catches a ball 3.2 s after throwing it vertically upward. (a) With what speed did he
throw it, and (b) what height did it reach?
Take upward to be
the positive direction.
Choose 𝑦
at the
height the ball was
thrown.
(a) Max Height
(a) Initial Speed
1
𝑦 + 𝑣 𝑡 − 𝑔𝑡 2
𝑦
2
𝑣
1
𝑦 − 𝑦 + 𝑔𝑡 2
2
𝑡
𝑣2
𝑦
𝑣
1
9.8 3.2
2
𝑦
𝑣
15.68 m/s
𝑦
𝑣 ≈ 16 m/s
𝑣
2
− 2𝑔 𝑦 − 𝑦
𝑦 +
𝑣
2
− 𝑣2
2𝑎
15.68 2 −
+
2 9.8
12.54 m
𝑦 ≈ 13 m
4. A helicopter is ascending vertically with a speed of 5.10 m/s. At a height of 105 m above the Earth, a
package is dropped from a window. How much time does it take for the packet to reach the ground?
Take downward to be
the positive direction.
Choose 𝑦
at the
height the package was
released such that
𝑦 1 5 m.
𝑦
1 5
1
𝑦 + 𝑣 𝑡 + 𝑔𝑡 2
2
−5.1𝑡 + 4.9𝑡 2
4.9𝑡 2 − 5.1𝑡 − 1 5
𝑡
𝑡
− −5.1 ±
−5.1 2 − 4 4.9 −1 5
2 4.9
5.18 s
5. A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an
altitude of 950 m. After this point, its acceleration is that of gravity, downward. (a) What is the velocity
of the rocket when it runs out of fuel? (b) How long does it take to reach this point? (c) What maximum
altitude does the rocket reach? (d) How much time (total) does it take to reach maximum altitude?
(a) Velocity when out of fuel
𝑣2
𝑣
𝑣
𝑣
+ 2𝑎 𝑦 − 𝑦
𝑣
𝑣 + 𝑎𝑡
+ 2 3.2 95
𝑡
𝑣−𝑣
𝑎
𝑡
24.37 s ≈ 24 s
2
77.97 m/s ≈ 78 m/s
(c) Max height
𝑣2
𝑦
𝑣
(b) Time to run out of fuel
2
𝑦 +
(d) Time to max height
− 2𝑔 𝑦 − 𝑦
𝑣
77.97 −
3.2
2
− 𝑣2
2𝑎
𝑦
77.97 2 −
95 +
2 9.8
𝑦
126 m
𝑣
𝑣 − 𝑔𝑡
𝑡
𝑣 −𝑣
𝑔
𝑡
7.96 s
77.97 −
9.8
Total Time:
𝑡
24.37 + 7.96
𝑡 ≈ 32 s
32.33 s
Name: _______________________
Mr. Kepple
Kinematics Problems 2
1D Kinematics – HW#7
Date: ___________ Period: _____
The problems on this page are more challenging than simply plugging into the kinematics equations.
You will have to do additional thinking and follow multiple logical steps to the find the solutions.
1. A falling stone takes 0.33 s to travel past a window 2.2 m tall.
From what height above the top of the window did the stone fall?
Speed at the top of the window, take downward to be
positive and 𝑦0 = 0 at the top of the window.
1
𝑦 = 𝑦0 + 𝑣0 𝑡 + 𝑔𝑡 2
2
1
𝑦 − 𝑦0 − 𝑔𝑡 2 2.2 − 0 − 4.9 0.33
2
𝑣0 =
=
𝑡
0.33
2
= 5.05 m/s
The height above the top of the window is the distance required to reach this
speed when dropped from rest.
0 − 5.05 2
𝑦0 = 0 −
𝑣 2 = 𝑣0 2 − 2𝑔 𝑦 − 𝑦0
2 9.8
𝑣0 2 − 𝑣 2
𝑦0 = 𝑦 −
2𝑔
𝑦0 = 1.3 m
2. A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4 s later. If the
speed of sound is 340 m/s, how high is the cliff?
total time
Combine equations and eliminate 𝑡:
𝑇 = 𝑡 + 𝑡𝑠
𝑡 = 𝑇 − 𝑡𝑠
rock
2ℎ = 𝑔 𝑇 − 𝑡𝑠
1
𝑦 = 𝑦0 + 𝑣0 𝑡 − 𝑔𝑡
2
2
0 = 2ℎ + 0 − 𝑔𝑡 2
sound
𝑦 = 𝑦0 + 𝑣0 𝑡 + 𝑎𝑡
2
ℎ = 0 + 𝑣𝑠 𝑡𝑠 + 0
𝑡𝑠 =
ℎ
𝑣𝑠
2
𝑔 2 2𝑇ℎ ℎ2
ℎ=
𝑇 −
+ 2
2
𝑣𝑠
𝑣𝑠
2ℎ = 𝑔𝑡 2
1
𝑔
ℎ
ℎ=
𝑇−
2
𝑣𝑠
2
2
𝑇𝑔
𝑔𝑇 2
ℎ −
ℎ+
=0
𝑣𝑠
2
2𝑣𝑠 2
𝑔
2
4.239 × 10−5 ℎ2 − 1.098ℎ + 56.64 = 0
From quadric formula:
ℎ = 52 m
3. A ball is dropped from the top of a 50.0-m high cliff. At the same time, a carefully aimed stone is
thrown straight up from the bottom of the cliff with a speed of 24.0 m/s. The stone and ball collide
part way up. How far above the base of the cliff does this happen?
ball
1
𝑦 = 𝑦0 + 𝑣0 𝑡 − 𝑔𝑡 2
Set equations equation to each other
and solve for time:
2
𝑦 = 50 + 0 − 4.9𝑡 2
50 − 4.9𝑡 2 = 24𝑡 − 4.9𝑡 2
50 = 24𝑡
stone
1
𝑦 = 𝑦0 + 𝑣0 𝑡 + 𝑎𝑡 2
𝑡=
2
𝑦 = 0 + 24𝑡 − 4.9𝑡 2
50
= 2.083 s
24
Use time to find the position:
𝑦 = 50 − 4.9 2.083
2
= 28.7 m
4. A rock is thrown vertically upward with a speed of 12.0 m/s. Exactly 1.00 s later, a ball is thrown up
vertically along the same path with a speed of 18.0 m/s. (a) At what time will they strike each other?
(b) At what height will the collision occur?
time
rock
(a) Time until collision
𝑡𝑏 = 𝑡 − 1
12𝑡 − 4.9𝑡 2 = 27.8𝑡 − 22.9 − 4.9𝑡 2
1
𝑦 = 𝑦0 + 𝑣0 𝑡 − 𝑔𝑡 2
12𝑡 = 27.8𝑡 − 22.9
2
𝑦 = 0 + 12𝑡 − 4.9𝑡 2
ball
15.8𝑡 = 22.9
𝑡 = 1.45 s
1
𝑦 = 𝑦0 + 𝑣0 𝑡 − 𝑔𝑡 2
2
𝑦 = 0 + 18𝑡𝑏 − 4.9𝑡𝑏 2
𝑦 = 18 𝑡 − 1 − 4.9 𝑡 − 1
2
(b) Height of collision
𝑦 = 18𝑡 − 18 − 4.9 𝑡 2 − 2𝑡 + 1
𝑦 = 12𝑡 − 4.9𝑡 2
𝑦 = 18𝑡 − 18 − 4.9𝑡 2 + 9.8𝑡 − 4.9
𝑦 = 12 1.45 − 4.9 1.45
𝑦 = 27.8𝑡 − 22.9 − 4.9𝑡 2
𝑦 = 7.10 m
2