part four:
for part four of this project, you will answer the following questions using the kinematic equations for uniformly accelerated motion.
1. determine the acceleration (in m/s²) of an object that ....
a. moves in a straight line with a constant speed of 20.0 m/s for 12.0 seconds
b. changes its velocity from 12.1 m/s to 23.5 m/s in 7.81 seconds
c. changes its velocity from 0.0 mi/hr to 60.0 mi/hr in 4.20 seconds
d. accelerates from 33.4 m/s to 18.9 m/s over a distance of 109 m
2. determine the magnitude of the displacement (in meters) of an object that ....
a. moves from beenthere to donethat (with an average speed of 28.0 m/s) and then back to beenthere (with an average speed of 28.0 m/s) if both the forward and the return trip take 46 minutes each.
b. moves at a constant speed of 8.30 m/s in a straight line for 15.0 seconds.
c. decelerates at a rate of -4.35 m/s/s from a speed of 38.1 m/s to a speed of 17.6 m/s
d. accelerates from rest at a rate of 3.67 m/s² for 12.1 seconds
e. is moving at 12.2 m/s and then accelerates at a rate of +1.88 m/s² for 17.0 seconds
3. the hare is sleeping at a location that is 1200 m from the finish line. the tortoise passes him at a steady speed of 5.0 cm/s. if the hare finally wakes up 6.5 hours later, then what minimum acceleration (assumed constant) must he have in order to pass the tortoise before the finish line.
4. a gold car moving at 12.0 m/s passes a green car while the green car is at rest at a stoplight. the green car immediately accelerates at a rate of +1.80 m/s/s for 11.0 seconds and then maintains a constant speed. after how much time (relative to the initial starting time) must the green car drive before catching up with the gold car.
5. ima rilla saari is cruising at 28.0 m/s down lake avenue and through the forest preserve. she notices a deer jump into the road at a location 62.0 m in front of her. ima first reacts to the event, then slams on her brakes and decelerates at -8.10 m/s², and ultimately stops a picometer in front of the frozen deer. what is imas reaction time? (i.e., how long did it take ima to react to the event prior to decelerating?)

Question

part four:
for part four of this project, you will answer the following questions using the kinematic equations for uniformly accelerated motion.
1. determine the acceleration (in m/s²) of an object that ....
   a. moves in a straight line with a constant speed of 20.0 m/s for 12.0 seconds
   b. changes its velocity from 12.1 m/s to 23.5 m/s in 7.81 seconds
   c. changes its velocity from 0.0 mi/hr to 60.0 mi/hr in 4.20 seconds
   d. accelerates from 33.4 m/s to 18.9 m/s over a distance of 109 m
2. determine the magnitude of the displacement (in meters) of an object that ....
   a. moves from beenthere to donethat (with an average speed of 28.0 m/s) and then back to beenthere (with an average speed of 28.0 m/s) if both the forward and the return trip take 46 minutes each.
   b. moves at a constant speed of 8.30 m/s in a straight line for 15.0 seconds.
   c. decelerates at a rate of -4.35 m/s/s from a speed of 38.1 m/s to a speed of 17.6 m/s
   d. accelerates from rest at a rate of 3.67 m/s² for 12.1 seconds
   e. is moving at 12.2 m/s and then accelerates at a rate of +1.88 m/s² for 17.0 seconds
3. the hare is sleeping at a location that is 1200 m from the finish line. the tortoise passes him at a steady speed of 5.0 cm/s. if the hare finally wakes up 6.5 hours later, then what minimum acceleration (assumed constant) must he have in order to pass the tortoise before the finish line.
4. a gold car moving at 12.0 m/s passes a green car while the green car is at rest at a stoplight. the green car immediately accelerates at a rate of +1.80 m/s/s for 11.0 seconds and then maintains a constant speed. after how much time (relative to the initial starting time) must the green car drive before catching up with the gold car.
5. ima rilla saari is cruising at 28.0 m/s down lake avenue and through the forest preserve. she notices a deer jump into the road at a location 62.0 m in front of her. ima first reacts to the event, then slams on her brakes and decelerates at -8.10 m/s², and ultimately stops a picometer in front of the frozen deer. what is imas reaction time? (i.e., how long did it take ima to react to the event prior to decelerating?)

Accepted Answer

### 1a
# Explanation:
## Step1: Recall acceleration definition
Acceleration $a = \frac{\Delta v}{\Delta t}$. For constant speed, $\Delta v = 0$.
$a=\frac{0}{12.0}=0$
# Answer:
$0\space m/s^2$

### 1b
# Explanation:
## Step1: Identify $\Delta v$ and $\Delta t$
$\Delta v = 23.5 - 12.1 = 11.4\space m/s$, $\Delta t = 7.81\space s$
## Step2: Calculate acceleration
$a=\frac{\Delta v}{\Delta t}=\frac{11.4}{7.81}\approx1.46\space m/s^2$
# Answer:
$\approx1.46\space m/s^2$

### 1c
# Explanation:
## Step1: Convert velocity to m/s
$60.0\space mi/hr=\frac{60\times1609.34}{3600}\approx26.82\space m/s$, $\Delta v = 26.82 - 0 = 26.82\space m/s$, $\Delta t = 4.20\space s$
## Step2: Calculate acceleration
$a=\frac{26.82}{4.20}\approx6.386\space m/s^2$
# Answer:
$\approx6.39\space m/s^2$ (rounded)

### 1d
# Explanation:
## Step1: Use kinematic equation $v_f^2 = v_i^2 + 2ax$
$v_i = 33.4\space m/s$, $v_f = 18.9\space m/s$, $x = 109\space m$
## Step2: Solve for $a$
$a=\frac{v_f^2 - v_i^2}{2x}=\frac{18.9^2 - 33.4^2}{2\times109}=\frac{357.21 - 1115.56}{218}=\frac{-758.35}{218}\approx - 3.48\space m/s^2$
# Answer:
$\approx - 3.48\space m/s^2$

### 2a
# Explanation:
## Step1: Displacement definition
Displacement is net change in position. Forward and back to start, so net displacement $= 0$.
# Answer:
$0\space m$

### 2b
# Explanation:
## Step1: Use $x = vt$ (constant speed, $a = 0$)
$v = 8.30\space m/s$, $t = 15.0\space s$
$x = 8.30\times15.0 = 124.5\space m$
# Answer:
$124.5\space m$

### 2c
# Explanation:
## Step1: Use $v_f^2 = v_i^2 + 2ax$ to find $x$
$v_i = 38.1\space m/s$, $v_f = 17.6\space m/s$, $a = - 4.35\space m/s^2$
## Step2: Solve for $x$
$x=\frac{v_f^2 - v_i^2}{2a}=\frac{17.6^2 - 38.1^2}{2\times(-4.35)}=\frac{309.76 - 1451.61}{-8.7}=\frac{-1141.85}{-8.7}\approx131.25\space m$
# Answer:
$\approx131\space m$ (rounded)

### 2d
# Explanation:
## Step1: Use $x = v_i t+\frac{1}{2}at^2$, $v_i = 0$ (from rest)
$a = 3.67\space m/s^2$, $t = 12.1\space s$
## Step2: Calculate $x$
$x = 0+\frac{1}{2}\times3.67\times(12.1)^2=\frac{1}{2}\times3.67\times146.41\approx267.5\space m$
# Answer:
$\approx268\space m$ (rounded)

### 2e
# Explanation:
## Step1: Use $x = v_i t+\frac{1}{2}at^2$
$v_i = 12.2\space m/s$, $a = 1.88\space m/s^2$, $t = 17.0\space s$
## Step2: Calculate $x$
$x = 12.2\times17+\frac{1}{2}\times1.88\times(17)^2=207.4 + 0.94\times289=207.4 + 271.66 = 479.06\space m$
# Answer:
$\approx479\space m$

### 3
# Explanation:
## Step1: Find distance tortoise travels in $6.5\space hr$
Convert $6.5\space hr$ to seconds: $6.5\times3600 = 23400\space s$, $v = 5.0\space cm/s = 0.05\space m/s$, distance $d_t = 0.05\times23400 = 1170\space m$
Remaining distance to finish: $1200 - 1170 = 30\space m$ (tortoise's remaining distance). Hare needs to cover $1200\space m$ in time $t$ (same as tortoise's remaining time? Wait, no: Hare wakes up, tortoise has already traveled $1170\space m$, so tortoise has $30\space m$ left, time for tortoise: $t_t=\frac{30}{0.05}=600\space s$. Hare starts from rest ($v_i = 0$), needs to cover $1200\space m$ in $t = 600\space s$? Wait, no: Wait, initial distance from finish: $1200\space m$. Tortoise speed $0.05\space m/s$, so total time tortoise takes to reach finish: $t_{total}=\frac{1200}{0.05}=24000\space s$. Hare sleeps for $6.5\space hr = 23400\space s$, so hare has $24000 - 23400 = 600\space s$ to reach finish.
## Step2: Use $x = \frac{1}{2}at^2$ (hare starts from rest, $v_i = 0$)
$x = 1200\space m$, $t = 600\space s$
$1200=\frac{1}{2}a(600)^2$
$a=\frac{2\times1200}{600^2}=\frac{2400}{360000}\approx0.00667\space m/s^2$
Wait, correction: Wait, when hare wakes up, tortoise is at $d = 0.05\times23400 = 1170\space m$ from start, so $30\space m$ from finish. So hare needs to cover $1200\space m$ (from his starting point) in the time tortoise takes to cover $30\space m$. Time for tortoise: $t=\frac{30}{0.05}=600\space s$. So hare's time is $600\space s$, starts from rest, so $x = \frac{1}{2}at^2$, $1200=\frac{1}{2}a(600)^2$, $a=\frac{2\times1200}{360000}=\frac{2400}{360000}=0.006666...\approx0.00667\space m/s^2$
# Answer:
$\approx0.00667\space m/s^2$

### 4
# Explanation:
## Step1: Motion of Green Car: first accelerate, then constant speed. Gold Car: constant speed $v = 12.0\space m/s$
Acceleration phase of Green Car: $a = 1.80\space m/s^2$, $t_1 = 11.0\space s$
Velocity after acceleration: $v = at_1 = 1.80\times11 = 19.8\space m/s$
Distance covered in acceleration: $x_1=\frac{1}{2}at_1^2=\frac{1}{2}\times1.80\times121 = 108.9\space m$
Distance covered by Gold Car in $t_1$: $x_{g1}=12\times11 = 132\space m$
## Step2: After acceleration, Green Car moves at $19.8\space m/s$, let time after acceleration be $t_2$
Distance for Green Car: $x_{g2}=12(t_1 + t_2)$
Distance for Green Car: $x_{gc}=108.9 + 19.8t_2$
Set equal: $12(11 + t_2)=108.9 + 19.8t_2$
$132 + 12t_2 = 108.9 + 19.8t_2$
$19.8t_2 - 12t_2 = 132 - 108.9$
$7.8t_2 = 23.1$
$t_2=\frac{23.1}{7.8}\approx2.96\space s$
## Step3: Total time
$t_{total}=11 + 2.96\approx13.96\space s$
# Answer:
$\approx14.0\space s$ (rounded)

### 5
# Explanation:
## Step1: Reaction time: distance traveled at constant speed $v = 28.0\space m/s$ in time $t_r$, then deceleration to stop (distance $x_d$)
Total distance: $x_{total}=62.0\space m$ (since stops a picometer in front, negligible, so total distance $62\space m$)
Deceleration: $a = - 8.10\space m/s^2$, final velocity $v_f = 0$
## Step2: Use $v_f^2 = v_i^2 + 2ax_d$ for deceleration phase
$0 = 28^2 + 2\times(-8.10)x_d$
$x_d=\frac{784}{16.2}\approx48.395\space m$
## Step3: Distance in reaction time
$x_r = 62 - 48.395 = 13.605\space m$
## Step4: Reaction time $t_r=\frac{x_r}{v}=\frac{13.605}{28.0}\approx0.486\space s$
# Answer:
$\approx0.486\space s$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

1a

Step1: Recall acceleration definition

Step1: Identify \(\Delta v\) and \(\Delta t\)

Step2: Calculate acceleration

Step1: Convert velocity to m/s

Step2: Calculate acceleration

Answer:

1b