1. a runner starts from rest and runs for 5s accelerating at a rate of 2m/s^2. what is the velocity of the…
1. a runner starts from rest and runs for 5s accelerating at a rate of 2m/s^2. what is the velocity of the runner after 5 s?\n2. how far did the runner in problem \1\ run during 5s?\n3. a car is traveling at 10 m/s and slows to 3 m/s at a rate of 2.5 m/s^s. how long did it take for the car to slow down?\n4. a boy rides his bike in a straight line for 2 minutes. he speeds up from 2 m/s to 30 m/s. what was his acceleration?\n5. if a person starts from rest, and walks for 15m accelerating at.5 m.s^s. how much time does it take to walk the 15m?\nsome additional problems to try:\n1) a woman walks a distance of 360m with an average speed of 1.2 m/s. how much time did it take her to walk this distance?\n2) starting from rest and moving in a straight line, a runner achieves a velocity of 7 m/s in a time of 2 sec. what is the acceleration of the runner?\n3) the velocity of a car decreases from 30m/s to 18 m/s in 4 seconds, what is the acceleration of the car while this takes place?\n4) a runner traveling with an initial velocity of 2 m/s accelerates at 2.5 m/s^2 for 5 seconds. what is the final velocity of the runner? how far has the runner gone?\n5) a plane lanes of a runway with a velocity of 300 m/s and slows to a stop. it takes 10 sec to stop.\n a) what was rate of acceleration of the plane?\n b) if the runway was 3,000m long, did the plane stop on or off the runway?
Answer
Explanation:
1.
Step1: Identify the kinematic - equation
We use the equation $v = v_0+at$, where $v_0 = 0$ (starts from rest), $a = 2m/s^2$, and $t = 5s$. $v=0 + 2\times5$
Step2: Calculate the velocity
$v = 10m/s$
Answer:
$10m/s$
Explanation:
2.
Step1: Select the kinematic - equation
We use the equation $x=v_0t+\frac{1}{2}at^2$. Since $v_0 = 0$, the equation simplifies to $x=\frac{1}{2}at^2$. Here, $a = 2m/s^2$ and $t = 5s$. $x=\frac{1}{2}\times2\times5^2$
Step2: Calculate the distance
$x = 25m$
Answer:
$25m$
Explanation:
3.
Step1: Identify the kinematic - equation
We use the equation $v = v_0+at$. We need to solve for $t$, so we can re - arrange it to $t=\frac{v - v_0}{a}$. Here, $v_0 = 10m/s$, $v = 3m/s$, and $a=- 2.5m/s^2$ (negative because it's decelerating). $t=\frac{3 - 10}{-2.5}$
Step2: Calculate the time
$t=\frac{-7}{-2.5}=2.8s$
Answer:
$2.8s$
Explanation:
4.
Step1: Convert time to seconds
$t = 2$ minutes $=2\times60s=120s$. Then use the equation $a=\frac{v - v_0}{t}$, where $v_0 = 2m/s$, $v = 30m/s$, and $t = 120s$. $a=\frac{30 - 2}{120}$
Step2: Calculate the acceleration
$a=\frac{28}{120}=\frac{7}{30}\approx0.23m/s^2$
Answer:
$\frac{7}{30}m/s^2\approx0.23m/s^2$
Explanation:
5.
Step1: Select the kinematic - equation
We use the equation $x = v_0t+\frac{1}{2}at^2$. Since $v_0 = 0$, the equation is $x=\frac{1}{2}at^2$. We need to solve for $t$, so $t=\sqrt{\frac{2x}{a}}$. Here, $x = 15m$ and $a = 0.5m/s^2$. $t=\sqrt{\frac{2\times15}{0.5}}$
Step2: Calculate the time
$t=\sqrt{60}\approx7.75s$
Answer:
$\sqrt{60}s\approx7.75s$
Explanation:
Additional 1:
Step1: Use the speed formula
We know that $v=\frac{d}{t}$, so $t=\frac{d}{v}$. Here, $d = 360m$ and $v = 1.2m/s$. $t=\frac{360}{1.2}$
Step2: Calculate the time
$t = 300s$
Answer:
$300s$
Explanation:
Additional 2:
Step1: Use the acceleration formula
We use the equation $a=\frac{v - v_0}{t}$. Since $v_0 = 0$, $v = 7m/s$, and $t = 2s$. $a=\frac{7 - 0}{2}$
Step2: Calculate the acceleration
$a = 3.5m/s^2$
Answer:
$3.5m/s^2$
Explanation:
Additional 3:
Step1: Use the acceleration formula
We use the equation $a=\frac{v - v_0}{t}$. Here, $v_0 = 30m/s$, $v = 18m/s$, and $t = 4s$. $a=\frac{18 - 30}{4}$
Step2: Calculate the acceleration
$a=\frac{-12}{4}=-3m/s^2$
Answer:
$-3m/s^2$
Explanation:
Additional 4:
Step1: Calculate the final velocity
Use the equation $v = v_0+at$. Here, $v_0 = 2m/s$, $a = 2.5m/s^2$, and $t = 5s$. So $v=2+2.5\times5=2 + 12.5 = 14.5m/s$.
Step2: Calculate the distance
Use the equation $x=v_0t+\frac{1}{2}at^2$. So $x=2\times5+\frac{1}{2}\times2.5\times5^2=10+\frac{1}{2}\times2.5\times25=10 + 31.25 = 41.25m$
Answer:
Final velocity: $14.5m/s$, Distance: $41.25m$
Explanation:
Additional 5a:
Step1: Use the acceleration formula
We use the equation $a=\frac{v - v_0}{t}$. Here, $v_0 = 300m/s$, $v = 0m/s$, and $t = 10s$. $a=\frac{0 - 300}{10}$
Step2: Calculate the acceleration
$a=-30m/s^2$
Answer:
$-30m/s^2$
Explanation:
Additional 5b:
Step1: Use the kinematic - equation
We use the equation $x=v_0t+\frac{1}{2}at^2$. Here, $v_0 = 300m/s$, $a=-30m/s^2$, and $t = 10s$. $x=300\times10+\frac{1}{2}\times(-30)\times10^2=3000-1500 = 1500m$
Step2: Compare with runway length
Since $1500m<3000m$, the plane stops on the runway.
Answer:
The plane stops on the runway.