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?
Part A
The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.
From
v 2 = ( v 0 ) 2 + 2 ay , \begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\
\end{aligned} v 2 = ( v 0 ) 2 + 2 ay , we have
v 2 = ( v 0 ) 2 + 2 ay v 2 − 2 ay = ( v 0 ) 2 v 0 = v 2 − 2 ay \begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\
\text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}}
\end{aligned} v 2 v 2 − 2 ay v 0 = ( v 0 ) 2 + 2 ay = ( v 0 ) 2 = v 2 − 2 ay Substituting the known values,
v 0 = v 2 − 2 ay v 0 = 0 2 − 2 ( − 9.81 m/s 2 ) ( 2.50 m ) v 0 = 7.00 m/s \begin{aligned}
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\
\text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\
\text{v}_0&= {\color{green}7.00 \ \text{m/s}}
\end{aligned} v 0 v 0 v 0 = v 2 − 2 ay = 0 2 − 2 ( − 9.81 m/s 2 ) ( 2.50 m ) = 7.00 m/s Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.
Part B
Since the motion of the kangaroo has uniform acceleration, we can use the formula
y = v o t + 1 2 a t 2 \text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2 y = v o t + 2 1 a t 2 The initial and final position of the kangaroo will be the same, so y y y 2 .
y = v 0 t + 1 2 a t 2 0 = ( 7.00 m/s ) t + 1 2 ( − 9.81 m/s 2 ) t 2 0 = 7 t − 4.905 t 2 7 t − 4.905 t 2 = 0 t ( 7 − 4.905 t ) = 0 t = 0 or 7 − 4.905 t = 0 \begin{aligned}
\text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\
0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\
0 & =7\text{t}-4.905\text{t}^{2}\\
7\text{t}-4.905\text{t}^{2}&=0 \\
\text{t}\left( 7-4.905\text{t} \right) & =0 \\
\text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\
\end{aligned} y 0 0 7 t − 4.905 t 2 t ( 7 − 4.905 t ) t = 0 = v 0 t + 2 1 a t 2 = ( 7.00 m/s ) t + 2 1 ( − 9.81 m/s 2 ) t 2 = 7 t − 4.905 t 2 = 0 = 0 or 7 − 4.905 t = 0 Discard the time 0 since this refers to the beginning of motion. Therefore, we have
7 − 4.905 t = 0 4.905 t = 7 t = 7 4.905 t = 1.43 s \begin{aligned}
7-4.905\text{t} &=0 \\
4.905\text{t} & = 7 \\
\text{t} & =\frac{7}{4.905} \\
\text{t}&={\color{green}1.43 \ \text{s}}
\end{aligned} 7 − 4.905 t 4.905 t t t = 0 = 7 = 4.905 7 = 1.43 s The kangaroo is about 1.43 seconds long in the air.
This entry was posted in Physics , Sciences and tagged College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Freefall , motion , motion of a kangaroo , Physics , Physics Solutions , Solution Manual for College Physics by Openstax on December 26, 2020 by Engineering Math .
If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road? SOLUTION:
Draw the free-body diagram of the car
Consider the vertical direction
Consider the motion in the horizontal direction
Solve for the acceleration of the car.
Solve for the coefficient of kinetic friction
This entry was posted in Engineering Mathematics Blog , Physics , Sciences and tagged College Physics , College Physics Solutions , Force and Motion , Friction , Kinematics , motion , motion in one dimension , Science , Solutions , university physics on December 29, 2018 by Engineering Math .
Freight trains can produce only relatively small accelerations and decelerations. (a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s? (b) If the train can slow down at a rate of 0.550 m/s2 , how long will it take to come to a stop from this velocity? (c) How far will it travel in each case? Solution:
Part A
We are given the the following: a = 0.0500 m/s 2 a=0.0500 \ \text{m/s}^2 a = 0.0500 m/s 2 t = 8.00 mins t=8.00 \ \text{mins} t = 8.00 mins v 0 = 4.00 m/s v_0=4.00 \ \text{m/s} v 0 = 4.00 m/s
The final velocity can be solved using the formula v f = v 0 + a t v_f=v_0+at v f = v 0 + a t
v f = v 0 + a t v f = 4.00 m/s + ( 0.0500 m/s 2 ) ( 8.00 min × 60 sec 1 min ) v f = 28.0 m/s ( Answer ) \begin{align*}
v_f& = v_0+at \\
v_f & = 4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right) \\
v_f & = 28.0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} v f v f v f = v 0 + a t = 4.00 m/s + ( 0.0500 m/s 2 ) ( 8.00 min × 1 min 60 sec ) = 28.0 m/s ( Answer ) Part B
Rearrange the equation we used in part (a) by solving in terms of t t t
t = v f − v 0 a t = 0 m/s − 28 m/s − 0.550 m/s 2 t = 50.91 sec ( Answer ) \begin{align*}
t & =\frac{{v_f}-v_0}{a} \\
t & = \frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2} \\
t & = 50.91\:\sec\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} t t t = a v f − v 0 = − 0.550 m/s 2 0 m/s − 28 m/s = 50.91 sec ( Answer ) Part C
The change in position for part (a), Δ x \Delta x Δ x Δ x = v 0 t + 1 2 a t 2 \Delta x=v_0 t+\frac{1}{2} at^2 Δ x = v 0 t + 2 1 a t 2
Δ x = v 0 t + 1 2 a t 2 Δ x = ( 4.0 m/s ) ( 480 s ) + 1 2 ( 0.0500 m/s 2 ) ( 480 s ) 2 Δ x = 7680 m ( Answer ) \begin{align*}
\Delta x & =v_0 t+\frac{1}{2} at^2 \\
\Delta x & =\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2 \\
\Delta x & = 7680\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} Δ x Δ x Δ x = v 0 t + 2 1 a t 2 = ( 4.0 m/s ) ( 480 s ) + 2 1 ( 0.0500 m/s 2 ) ( 480 s ) 2 = 7680 m ( Answer ) For the situation in part (b), the distance traveled is computed using the formula Δ x = v f 2 − v 0 2 2 a \Delta x=\frac{v_f^2-v_0^2}{2 a} Δ x = 2 a v f 2 − v 0 2
Δ x = ( 0 m/s ) 2 − ( 28.0 m/s ) 2 2 ( − 0.550 m/s 2 ) Δ x = 712.73 m ( Answer ) \begin{align*}
\Delta x & =\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)} \\
\Delta x & =712.73\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} Δ x Δ x = 2 ( − 0.550 m/s 2 ) ( 0 m/s ) 2 − ( 28.0 m/s ) 2 = 712.73 m ( Answer )
This entry was posted in Physics , Sciences and tagged acceleration , College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Kinematics , Mechanics , motion , one-dimensional kinematics , One-Dimensional Kinematics with Constant Acceleration , Physics Solution Manual , Physics Solutions , Science , Solution Manual for College Physics by Openstax , Speed , Velocity on December 18, 2016 by Engineering Math .
A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s.
(a) What is its average acceleration?
(b) How far does it travel in that time?
Solution:
We are given the following: v 0 = 0 m/s v_0=0 \ \text{m/s} v 0 = 0 m/s v f = 26.8 m/s v_f=26.8 \ \text{m/s} v f = 26.8 m/s t = 3.90 s t=3.90\ \text{s} t = 3.90 s
Part A
The average acceleration of the motorcycle can be solved using the equation a ‾ = Δ v Δ t \overline{a}=\frac{\Delta v}{\Delta t} a = Δ t Δ v
a ‾ = Δ v Δ t a ‾ = 26.8 m/s − 0 m/s 3.90 s a ‾ = 6.872 m/s 2 ( Answer ) \begin{align*}
\overline{a} & =\frac{\Delta v}{\Delta t} \\
\overline{a} & =\frac{26.8\:\text{m/s}-0\:\text{m/s}}{3.90\:\text{s}} \\
\overline{a} & =6.872\:\text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} a a a = Δ t Δ v = 3.90 s 26.8 m/s − 0 m/s = 6.872 m/s 2 ( Answer ) Part B
The distance traveled is equal to the average velocity multiplied by the time of travel. That is,
Δ x = v a v e t Δ x = ( 0 m/s + 26.8 m/s 2 ) ( 3.90 s ) Δ x = 52.26 m ( Answer ) \begin{align*}
\Delta x & =v_{ave}t\\
\Delta x & =\left(\frac{0\:\text{m/s}+26.8\:\text{m/s}}{2}\right)\left(3.90\:\text{s}\right) \\
\Delta x & =52.26\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*} Δ x Δ x Δ x = v a v e t = ( 2 0 m/s + 26.8 m/s ) ( 3.90 s ) = 52.26 m ( Answer )
This entry was posted in Physics , Sciences and tagged acceleration , College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Kinematics , motion , motion in one dimension , one-dimensional kinematics , One-Dimensional Kinematics with Constant Acceleration , Physics Solution Manual , Physics Solutions , Solution Manual for College Physics by Openstax , Speed , Time , Velocity on September 9, 2016 by Engineering Math .
In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33× 10-2 s, calculate the distance over which the puck accelerates. Solution:
The best equation that can be used to solve this problem is Δ x = v a v e t \Delta x=v_{ave} t Δ x = v a v e t
Δ x = v a v e t Δ x = ( 8 m/s + 40 m/s 2 ) ( 3.33 × 1 0 − 2 s ) Δ x = 0.7992 m ( Answer ) \begin{align*}
\Delta x & = v_{ave} t \\
\Delta x & = \left(\frac{8\:\text{m/s}+40\:\text{m/s}}{2}\right)\left(3.33\times 10^{-2}\:\text{s}\right) \\
\Delta x & = 0.7992\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} Δ x Δ x Δ x = v a v e t = ( 2 8 m/s + 40 m/s ) ( 3.33 × 1 0 − 2 s ) = 0.7992 m ( Answer ) Therefore, the distance over which the puck accelerates is 0.7992 meters.
This entry was posted in Physics , Sciences and tagged acceleration , College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Kinematics , motion , motion in one dimension , one-dimensional kinematics , One-Dimensional Kinematics with Constant Acceleration , Physics Solution Manual , Physics Solutions , Solution Manual for College Physics by Openstax , Speed , Time , Velocity on September 9, 2016 by Engineering Math .
Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart.
(a) Make a sketch of the solution.
(b) List the knowns in this problem.
(c) How long does the acceleration take? To solve this part, identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units.
(d) Is the answer reasonable when compared with the time for a heartbeat?
Solution:
Part A
The sketch should contain the starting point and the final point. This will be done by connecting a straight line from the starting point to the final point. The sketch is shown below.
Part B
The list of known variables are:
Initial velocity: v 0 = 0 m/s v_0=0\:\text{m/s} v 0 = 0 m/s v f = 30.0 cm/s v_f=30.0\:\text{cm/s} v f = 30.0 cm/s x − x 0 = 1.80 cm x-x_0=1.80\:\text{cm} x − x 0 = 1.80 cm
Part C
The best equation to solve for this is Δ x = v ave t \Delta \text{x}=\text{v}_{\text{ave}}\text{t} Δ x = v ave t v a v e v_{ave} v a v e t t t
Δ x = v a v e t t = Δ x v a v e t = 1.80 cm ( 0 cm/s + 30 cm/s ) 2 t = 0.12 s ( Answer ) \begin{align*}
\Delta x & =v_{ave} t \\
t &=\frac{\Delta x}{v_{ave}} \\
t & =\frac{1.80\:\text{cm}}{\frac{\left(0\:\text{cm/s}+30\:\text{cm/s}\right)}{2}}\\
t & =0.12\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*} Δ x t t t = v a v e t = v a v e Δ x = 2 ( 0 cm/s + 30 cm/s ) 1.80 cm = 0.12 s ( Answer ) Part D
Since the computed value of the time for the acceleration of blood out of the ventricle is only 0.12 seconds (only a fraction of a second), the answer seems reasonable. This is due to the fact that an entire heartbeat cycle takes about one second. So, the answer is yes, the answer is reasonable.
This entry was posted in Physics , Sciences and tagged acceleration , College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Kinematics , motion , motion in one dimension , one-dimensional kinematics , One-Dimensional Kinematics with Constant Acceleration , Physics Solution Manual , Physics Solutions , Solution Manual for College Physics by Openstax , Speed , Time , Velocity on September 8, 2016 by Engineering Math .
At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s2 . (a) How far does she travel in the next 5.00 s? (b) What is her final velocity? (c) Evaluate the result. Does it make sense? Solution:
We are given the following: v 0 = 9.00 m/s v_0=9.00 \ \text{m/s} v 0 = 9.00 m/s a = 2.00 m/s 2 a=2.00 \ \text{m/s}^2 a = 2.00 m/s 2
Part A
For this part, we are given t = 5.00 s t=5.00 \ \text{s} t = 5.00 s x = x 0 + v 0 t + 1 2 a t 2 x=x_0+v_0 t+\frac{1}{2}at^2 x = x 0 + v 0 t + 2 1 a t 2
x = x 0 + v 0 t + 1 2 a t 2 x = 0 m + ( 9.00 m/s ) ( 5.00 s ) + 1 2 ( − 2.00 m/s 2 ) ( 5.00 s ) 2 x = 20 meters ( Answer ) \begin{align*}
x & =x_0+v_0 t+\frac{1}{2}at^2 \\
x & =0\:\text{m}+\left(9.00\:\text{m/s}\right)\left(5.00\:\text{s}\right)+\frac{1}{2}\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right)^2 \\
x & =20\:\text{meters} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*} x x x = x 0 + v 0 t + 2 1 a t 2 = 0 m + ( 9.00 m/s ) ( 5.00 s ) + 2 1 ( − 2.00 m/s 2 ) ( 5.00 s ) 2 = 20 meters ( Answer ) Part B
The final velocity can be determined using the formula v f = v 0 + a t v_f=v_0+at v f = v 0 + a t
v f = v 0 + a t v f = 9.00 m/s + ( − 2.00 m/s 2 ) ( 5.00 s ) v f = − 1 m/s ( Answer ) \begin{align*}
v_f & =v_0+at \\
v_f & =9.00\:\text{m/s}+\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right) \\
v_f & =-1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*} v f v f v f = v 0 + a t = 9.00 m/s + ( − 2.00 m/s 2 ) ( 5.00 s ) = − 1 m/s ( Answer ) Part C
The result says that the runner starts at the rate of 9 m/s and decelerates at 2 m/s2 . After some time, the velocity is already negative. This does not make sense because if the velocity is negative, that means that the runner is already running backwards.
This entry was posted in Physics , Sciences and tagged acceleration , College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Kinematics , motion , motion in one dimension , one-dimensional kinematics , One-Dimensional Kinematics with Constant Acceleration , Physics Solution Manual , Physics Solutions , Solution Manual for College Physics by Openstax , Speed , Time , Velocity on September 7, 2016 by Engineering Math .
A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.2 0×105 m/s2 for 8.10× 10-4 s . What is its muzzle velocity (that is, its final velocity)? Solution:
We are given the following: a = 6.20 × 1 0 5 m/s 2 a=6.20 \times 10^{5} \ \text{m/s}^2 a = 6.20 × 1 0 5 m/s 2 Δ t = 8.10 × 1 0 − 4 s \Delta t=8.10 \times 10^{-4} \ \text{s} Δ t = 8.10 × 1 0 − 4 s v 0 = 0 m/s v_0=0 \text{m/s} v 0 = 0 m/s
The muzzle velocity of the bullet is computed as follows:
v f = v 0 + a t v f = 0 m/s + ( 6.20 × 1 0 5 m/s 2 ) ( 8.10 × 1 0 − 4 s ) v f = 502 m/s ( Answer ) \begin{align*}
v_f & =v_0+at \\
v_f & = 0\:\text{m/s}+\left(6.20\times 10^5\text{ m/s}^2\right)\left(8.10\times 10^{-4}\:\text{s}\right) \\
v_f & =502\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} v f v f v f = v 0 + a t = 0 m/s + ( 6.20 × 1 0 5 m/s 2 ) ( 8.10 × 1 0 − 4 s ) = 502 m/s ( Answer ) Therefore, the muzzle velocity, or final velocity, is 502 m/s.
This entry was posted in Physics , Sciences and tagged acceleration , College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Kinematics , Kinematics in Two Dimensions , kinematics of Rotational Motion , motion , one-dimensional kinematics , One-Dimensional Kinematics with Constant Acceleration , Physics Solution Manual , Physics Solutions , Solution Manual for College Physics by Openstax , two-dimensional kinematics on August 26, 2016 by Engineering Math .
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10× 104 m/s2 , and 1.85 ms (1 ms = 10-3 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball? Solution:
We are given the following values: a = − 2.10 × 1 0 4 m/s 2 ; t = 1.85 × 1 0 − 3 s ; v f = 0 m/s a=-2.10 \times 10^4 \ \text{m/s}^2; t=1.85 \times 10^{-3} \ \text{s}; v_f=0 \ \text{m/s} a = − 2.10 × 1 0 4 m/s 2 ; t = 1.85 × 1 0 − 3 s ; v f = 0 m/s
The formula in solving for the initial velocity is
v 0 = v f − a t v_0=v_f-at v 0 = v f − a t Substitute the given values
v 0 = 0 m/s − ( − 2.10 × 1 0 4 m/s 2 ) ( 1.85 × 1 0 − 3 s ) v 0 = 38.85 m/s ( Answer ) \begin{align*}
v_0 & =0\:\text{m/s}-\left(-2.10\times 10^4\text{ m/s}^2\right)\left(1.85\times 10^{-3}\:\text{s}\right) \\
v_0 & =38.85\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} v 0 v 0 = 0 m/s − ( − 2.10 × 1 0 4 m/s 2 ) ( 1.85 × 1 0 − 3 s ) = 38.85 m/s ( Answer )
This entry was posted in Physics , Sciences and tagged acceleration , College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Kinematics , Kinematics in Two Dimensions , kinematics of Rotational Motion , motion , one-dimensional kinematics , One-Dimensional Kinematics with Constant Acceleration , Physics Solution Manual , Physics Solutions , Solution Manual for College Physics by Openstax , two-dimensional kinematics on August 25, 2016 by Engineering Math .
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