Question
2.5 Constant acceleration
1.An Olympic-cass sprinter starts a race with an acceleration of 4.50 m/s2
. (a)What is
her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.
2.A well-thrown bal is caught in a wel-padded mitt. If the deceleration of the ball is
2.10×104 m/s2
, and 1.85 ms(1 ms = 103 s)elapses from the time the ball first touches
the mitt until it stops, what was theinitia velocity of the ball?
3.A bullet in a gun is accelerated from the firing chamber to the end of the barre at an
averagerate of6.20×105 m/s
2 for8.10×10-4 s. What is its muzzle velocity (that is, its
final velocity)?
4.A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90
s. (a) whatis its average acceleration? (b) How far does it travel in that time?
5.A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of
0.250 m. (a) How long did the acceleration last? (b) Calculate the acceleration.
2.7 Falling object
1.Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and
(d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the
point of release to bey0 =0.
2.Abasketball referee tosses the ball straight up for the starting tip-off. At what
velocity must a basketball player eave the ground to rise 1.25 m above the floor in
an attempt to get the ball?
3.Calculate the height of a cliff if it takes 2.35 s for arock to hit the ground when it is
thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) howlong
would it take to reach the ground if it is thrown straight down with the same speed?
4.A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed
when it leaves the ground. (b) How long is it in the air?
5.A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at
10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position
and velocity 4.00 s after being released, and (c) the time before it hits the ground.
2.8 Graphical analysis
1.A graph ofv(t)is shown for a world-class track sprinter in a 100-m race. (See Figure
2.65). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous
velocity att= 5s? (c) What is his average acceleration between 0 and 4 s? (d) What is
his time for the race?
1.An Olympic-cass sprinter starts a race with an acceleration of 4.50 m/s2
. (a)What is
her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.
2.A well-thrown bal is caught in a wel-padded mitt. If the deceleration of the ball is
2.10×104 m/s2
, and 1.85 ms(1 ms = 103 s)elapses from the time the ball first touches
the mitt until it stops, what was theinitia velocity of the ball?
3.A bullet in a gun is accelerated from the firing chamber to the end of the barre at an
averagerate of6.20×105 m/s
2 for8.10×10-4 s. What is its muzzle velocity (that is, its
final velocity)?
4.A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90
s. (a) whatis its average acceleration? (b) How far does it travel in that time?
5.A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of
0.250 m. (a) How long did the acceleration last? (b) Calculate the acceleration.
2.7 Falling object
1.Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and
(d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the
point of release to bey0 =0.
2.Abasketball referee tosses the ball straight up for the starting tip-off. At what
velocity must a basketball player eave the ground to rise 1.25 m above the floor in
an attempt to get the ball?
3.Calculate the height of a cliff if it takes 2.35 s for arock to hit the ground when it is
thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) howlong
would it take to reach the ground if it is thrown straight down with the same speed?
4.A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed
when it leaves the ground. (b) How long is it in the air?
5.A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at
10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position
and velocity 4.00 s after being released, and (c) the time before it hits the ground.
2.8 Graphical analysis
1.A graph ofv(t)is shown for a world-class track sprinter in a 100-m race. (See Figure
2.65). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous
velocity att= 5s? (c) What is his average acceleration between 0 and 4 s? (d) What is
his time for the race?
Answers
Answered by
GPT-4o mini
Let's solve each part of your questions step by step.
### Constant Acceleration Problems
1. **Olympic-class sprinter**
- (a) To find the speed after 2.40 s of acceleration \( a = 4.50 \, \text{m/s}^2 \):
\[
v = u + at
\]
where \( u = 0 \, \text{m/s} \) (starting from rest), \( a = 4.50 \, \text{m/s}^2 \), and \( t = 2.40 \, \text{s} \):
\[
v = 0 + (4.50)(2.40) = 10.80 \, \text{m/s}
\]
- (b) For the position vs. time, the displacement can be calculated using:
\[
s = ut + \frac{1}{2} at^2
\]
Since \( u = 0 \):
\[
s = 0 + \frac{1}{2} (4.50)(2.40^2) = \frac{1}{2} (4.50)(5.76) = 12.96 \, \text{m}
\]
(Graph sketch: a parabolic curve that starts at the origin and reaches \( (2.40, 12.96) \)).
2. **Ball caught in a mitt**
- Given \( a = -2.10 \times 10^4 \, \text{m/s}^2 \) and \( t = 1.85 \, \text{ms} = 0.00185 \, \text{s} \):
Use:
\[
v = u + at
\]
Rearranging for initial velocity \( u \):
\[
u = v - at \quad (\text{final velocity } v = 0)
\]
\[
u = 0 - (-2.10 \times 10^4)(0.00185) = 38.85 \, \text{m/s}
\]
3. **Bullet acceleration**
- Given \( a = 6.20 \times 10^5 \, \text{m/s}^2 \) and \( t = 8.10 \times 10^{-4} \, \text{s} \):
\[
v = u + at \quad (u = 0)
\]
\[
v = 0 + (6.20 \times 10^5)(8.10 \times 10^{-4}) = 503.4 \, \text{m/s}
\]
4. **Motorcycle acceleration**
- (a) Average acceleration \( a = \frac{\Delta v}{\Delta t} = \frac{26.8 - 0}{3.90} = 6.87 \, \text{m/s}^2 \)
- (b) Distance traveled:
\[
s = ut + \frac{1}{2} at^2 = 0 + \frac{1}{2}(6.87)(3.90^2) = 53.2 \, \text{m}
\]
5. **Fireworks shell**
- (a) To find time \( t \):
\[
v^2 = u^2 + 2as \quad (u = 0)
\]
\[
65.0^2 = 0 + 2a(0.250) \implies a = \frac{65^2}{2 \cdot 0.250} = 8450 \, \text{m/s}^2
\]
- (b) Time using \( v = at \):
\[
t = \frac{v}{a} = \frac{65.0}{8450} \approx 0.0077 \, \text{s}
\]
### Falling Object Problems
1. **Ball thrown straight up (height and velocity)**
- Use \( y(t) = v_0 t - \frac{1}{2} g t^2 \) where \( g = 9.81 \, \text{m/s}^2 \) and \( v_0 = 15.0 \, \text{m/s} \)
- (a) \( t = 0.5 \, \text{s}\):
\[
y(0.5) = 15.0(0.5) - 0.5(9.81)(0.5^2) = 7.50 - 1.23 \approx 6.27 \, \text{m}
\]
- (b) \( t = 1.0 \, \text{s}\):
\[
y(1.0) = 15.0(1.0) - 0.5(9.81)(1.0^2) = 15.00 - 4.905 \approx 10.095 \, \text{m}
\]
- (c) \( t = 1.5 \, \text{s}\):
\[
y(1.5) = 15.0(1.5) - 0.5(9.81)(1.5^2) = 22.50 - 11.02375 \approx 11.47625 \, \text{m}
\]
- (d) \( t = 2.0 \, \text{s}\):
\[
y(2.0) = 15.0(2.0) - 0.5(9.81)(2.0^2) = 30.00 - 19.62 \approx 10.38 \, \text{m}
\]
2. **Basketball player’s jump**
- Using the formula \( v^2 = u^2 + 2as \) where \( s = 1.25 \) m and \( a = -9.81 \, \text{m/s}^2\),
\[
0 = u^2 + 2(-9.81)(1.25) \implies u^2 = 24.525 \implies u = \sqrt{24.525} \approx 4.95 \, \text{m/s}
\]
3. **Height of a cliff**
- (a) Using \( s = v_0 t - \frac{1}{2} g t^2 \):
\(
s = 8.00(2.35) - 0.5(9.81)(2.35^2) = 18.8 - 27.04 \approx -8.24 \text{m (below initial launch point)}
\) so height of cliff = 8.24 m.
- (b) If thrown down, use \( s = v_0 t + \frac{1}{2} g t^2 \):
\[
s = v_0 t + \frac{1}{2} g t^2 = 8.00(2.35) + 0.5(9.81)(2.35^2) \approx 18.8 + 27.04 \approx 45.84 \, \text{m}
\]
4. **Kangaroo jump**
- (a) Using \( v^2 = 2gs \):
\[
v^2 = 2(9.81)(2.50) \implies v \approx 7.0 \, \text{m/s}
\]
- (b) Use \( v = u + at \):
\[
t = \frac{v}{g} \implies t = \frac{7.0}{9.81} \approx 0.71 \, \text{s}
\]
5. **Coin dropped from balloon**
- (a) Rise height first: \( 10.0 \text{m/s} \times 0.500 \text{s} + \frac{1}{2} (9.81)(0.500^2) \)
\[
= 5 + 1.225 = 6.225 \text{m}
\]
Max height = \( 300 + 6.225 = 306.225 \) m
- (b) At \( t = 4.0 \text{s} \):
\[
y = 300 + 10(4) - \frac{1}{2}(9.8)(4^2) \approx 300 + 40 - 78.4 \approx 261.6 \, \text{m}
\]
- (c) Time to hit ground:
\[
0 = 300 + 10t - \frac{1}{2}(9.81)t^2 \quad \text{gives a quadratic equation}
\]
### Graphical Analysis Problems
1. Without the provided graph, I cannot calculate the values for the sprinter questions. If you can describe or provide the details of the graph, I can help with calculating average velocity, instantaneous velocity, average acceleration, and race time.
If you have specific values for these questions to clarify or wish to focus on particular problems, let me know!
### Constant Acceleration Problems
1. **Olympic-class sprinter**
- (a) To find the speed after 2.40 s of acceleration \( a = 4.50 \, \text{m/s}^2 \):
\[
v = u + at
\]
where \( u = 0 \, \text{m/s} \) (starting from rest), \( a = 4.50 \, \text{m/s}^2 \), and \( t = 2.40 \, \text{s} \):
\[
v = 0 + (4.50)(2.40) = 10.80 \, \text{m/s}
\]
- (b) For the position vs. time, the displacement can be calculated using:
\[
s = ut + \frac{1}{2} at^2
\]
Since \( u = 0 \):
\[
s = 0 + \frac{1}{2} (4.50)(2.40^2) = \frac{1}{2} (4.50)(5.76) = 12.96 \, \text{m}
\]
(Graph sketch: a parabolic curve that starts at the origin and reaches \( (2.40, 12.96) \)).
2. **Ball caught in a mitt**
- Given \( a = -2.10 \times 10^4 \, \text{m/s}^2 \) and \( t = 1.85 \, \text{ms} = 0.00185 \, \text{s} \):
Use:
\[
v = u + at
\]
Rearranging for initial velocity \( u \):
\[
u = v - at \quad (\text{final velocity } v = 0)
\]
\[
u = 0 - (-2.10 \times 10^4)(0.00185) = 38.85 \, \text{m/s}
\]
3. **Bullet acceleration**
- Given \( a = 6.20 \times 10^5 \, \text{m/s}^2 \) and \( t = 8.10 \times 10^{-4} \, \text{s} \):
\[
v = u + at \quad (u = 0)
\]
\[
v = 0 + (6.20 \times 10^5)(8.10 \times 10^{-4}) = 503.4 \, \text{m/s}
\]
4. **Motorcycle acceleration**
- (a) Average acceleration \( a = \frac{\Delta v}{\Delta t} = \frac{26.8 - 0}{3.90} = 6.87 \, \text{m/s}^2 \)
- (b) Distance traveled:
\[
s = ut + \frac{1}{2} at^2 = 0 + \frac{1}{2}(6.87)(3.90^2) = 53.2 \, \text{m}
\]
5. **Fireworks shell**
- (a) To find time \( t \):
\[
v^2 = u^2 + 2as \quad (u = 0)
\]
\[
65.0^2 = 0 + 2a(0.250) \implies a = \frac{65^2}{2 \cdot 0.250} = 8450 \, \text{m/s}^2
\]
- (b) Time using \( v = at \):
\[
t = \frac{v}{a} = \frac{65.0}{8450} \approx 0.0077 \, \text{s}
\]
### Falling Object Problems
1. **Ball thrown straight up (height and velocity)**
- Use \( y(t) = v_0 t - \frac{1}{2} g t^2 \) where \( g = 9.81 \, \text{m/s}^2 \) and \( v_0 = 15.0 \, \text{m/s} \)
- (a) \( t = 0.5 \, \text{s}\):
\[
y(0.5) = 15.0(0.5) - 0.5(9.81)(0.5^2) = 7.50 - 1.23 \approx 6.27 \, \text{m}
\]
- (b) \( t = 1.0 \, \text{s}\):
\[
y(1.0) = 15.0(1.0) - 0.5(9.81)(1.0^2) = 15.00 - 4.905 \approx 10.095 \, \text{m}
\]
- (c) \( t = 1.5 \, \text{s}\):
\[
y(1.5) = 15.0(1.5) - 0.5(9.81)(1.5^2) = 22.50 - 11.02375 \approx 11.47625 \, \text{m}
\]
- (d) \( t = 2.0 \, \text{s}\):
\[
y(2.0) = 15.0(2.0) - 0.5(9.81)(2.0^2) = 30.00 - 19.62 \approx 10.38 \, \text{m}
\]
2. **Basketball player’s jump**
- Using the formula \( v^2 = u^2 + 2as \) where \( s = 1.25 \) m and \( a = -9.81 \, \text{m/s}^2\),
\[
0 = u^2 + 2(-9.81)(1.25) \implies u^2 = 24.525 \implies u = \sqrt{24.525} \approx 4.95 \, \text{m/s}
\]
3. **Height of a cliff**
- (a) Using \( s = v_0 t - \frac{1}{2} g t^2 \):
\(
s = 8.00(2.35) - 0.5(9.81)(2.35^2) = 18.8 - 27.04 \approx -8.24 \text{m (below initial launch point)}
\) so height of cliff = 8.24 m.
- (b) If thrown down, use \( s = v_0 t + \frac{1}{2} g t^2 \):
\[
s = v_0 t + \frac{1}{2} g t^2 = 8.00(2.35) + 0.5(9.81)(2.35^2) \approx 18.8 + 27.04 \approx 45.84 \, \text{m}
\]
4. **Kangaroo jump**
- (a) Using \( v^2 = 2gs \):
\[
v^2 = 2(9.81)(2.50) \implies v \approx 7.0 \, \text{m/s}
\]
- (b) Use \( v = u + at \):
\[
t = \frac{v}{g} \implies t = \frac{7.0}{9.81} \approx 0.71 \, \text{s}
\]
5. **Coin dropped from balloon**
- (a) Rise height first: \( 10.0 \text{m/s} \times 0.500 \text{s} + \frac{1}{2} (9.81)(0.500^2) \)
\[
= 5 + 1.225 = 6.225 \text{m}
\]
Max height = \( 300 + 6.225 = 306.225 \) m
- (b) At \( t = 4.0 \text{s} \):
\[
y = 300 + 10(4) - \frac{1}{2}(9.8)(4^2) \approx 300 + 40 - 78.4 \approx 261.6 \, \text{m}
\]
- (c) Time to hit ground:
\[
0 = 300 + 10t - \frac{1}{2}(9.81)t^2 \quad \text{gives a quadratic equation}
\]
### Graphical Analysis Problems
1. Without the provided graph, I cannot calculate the values for the sprinter questions. If you can describe or provide the details of the graph, I can help with calculating average velocity, instantaneous velocity, average acceleration, and race time.
If you have specific values for these questions to clarify or wish to focus on particular problems, let me know!