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A driver gets on mile 25 of a highway at 3:30 pm and exits at mile 150 at 5:30 pm. If the road is straight, what is the velocity and is it average or instantaneous?
Solution:
v
average
=
Δ
x
Δ
t
=
x
f
−
x
i
t
f
−
t
i
=
150
−
25
5.5
−
3.5
=
125
2
=
62.5
miles
hour
{\displaystyle v_{\text{average}}={\frac {\Delta x}{\Delta t}}={\frac {x_{f}-x_{i}}{t_{f}-t_{i}}}={\frac {150-25}{5.5-3.5}}={\frac {125}{2}}=62.5{\frac {\text{miles}}{\text{hour}}}}
It is an average velocity because the time interval is not infinitesimally small. (In physics we like to be precise and call it velocity and not speed because if the person was going in the opposite direction, the result would have been negative. Velocity has direction as a property, speed does not.
A particle starting 23 m from the origin has moved to 43 m in 5 s. Find
v
¯
{\displaystyle {\bar {v}}}
.
Solution:
v
¯
=
x
f
−
x
i
t
f
−
t
i
=
43
m
−
23
m
5
s
−
0
s
=
20
m
5
s
=
4
m
/
s
{\displaystyle {\bar {v}}={\frac {x_{f}-x_{i}}{t_{f}-t_{i}}}={\frac {43\ \mathrm {m} -23\ \mathrm {m} }{5\ \mathrm {s} -0\ \mathrm {s} }}={\frac {20\ \mathrm {m} }{5\ \mathrm {s} }}=4\ \mathrm {m} /\mathrm {s} }
Finding the average acceleration if the direction reverses in the time interval [ edit | edit source ]
A person is jogging east at 3m/s when he suddenly reverses direction and is jogging west at 3m/s, taking one second to accomplish this reversal. Take east to be the 'positive' direction. What is the average acceleration?
Solution:
a
=
v
2
−
v
1
t
2
−
t
1
=
3
m
/
s
−
−
3
m
/
s
1
s
=
6
m
/
s
2
{\displaystyle a={\frac {v_{2}-v_{1}}{t_{2}-t_{1}}}={\frac {3m/s--3m/s}{1s}}=6m/s^{2}}
Wait a minute! East is postive and west is negative. There is a minus sign error
Use algebra to show that
v
=
v
0
+
a
t
{\displaystyle v=v_{0}+at}
and
x
=
x
0
+
v
0
t
+
1
2
a
t
2
{\displaystyle x=x_{0}+v_{0}t+{\tfrac {1}{2}}at^{2}}
implies
v
2
=
v
0
2
+
2
a
(
x
−
x
0
)
{\displaystyle v^{2}=v_{0}^{2}+2a\left(x-x_{0}\right)}
and
x
−
x
0
=
v
0
+
v
2
t
{\displaystyle x-x_{0}={\frac {v_{0}+v}{2}}t}
.
CALCULUS: Motion and the mathematical definition of derivative as a limit [ edit | edit source ]
By definition, velocity involves two different positions at two different times. However, we may take the limit that these differences are very small and define the instantaneous velocity .
v
(
t
)
=
lim
Δ
t
→
0
Δ
x
Δ
t
=
lim
Δ
t
→
0
x
(
t
f
)
−
x
(
t
i
)
t
f
−
t
i
{\displaystyle v(t)=\lim _{\Delta t\to 0}{\frac {\Delta x}{\Delta t}}=\lim _{\Delta t\to 0}{\frac {x(t_{f})-x(t_{i})}{t_{f}-t_{i}}}}
A connection to differential calculus is seen by rewriting
t
i
{\displaystyle t_{i}}
and
t
f
{\displaystyle t_{f}}
as
t
{\displaystyle t}
and
t
+
Δ
t
{\displaystyle t+\Delta t}
, so that
x
f
=
x
(
t
+
Δ
t
)
{\displaystyle x_{f}=x(t+\Delta t)}
:
v
(
t
)
=
lim
Δ
t
→
0
x
(
t
+
Δ
t
)
−
x
(
t
)
Δ
t
=
d
x
d
t
{\displaystyle v(t)=\lim _{\Delta t\to 0}{\frac {x(t+\Delta t)-x(t)}{\Delta t}}={\frac {dx}{dt}}}
CALCULUS: Problem involving velocity, acceleration, and equations of motion[ edit | edit source ]
A particle's motion[ 1]
is described by the equation
x
(
t
)
=
2
t
3
+
5
t
+
2
{\displaystyle \ x(t)=2t^{3}+5t+2}
. Find
a) the particle's velocity function,
b) its instantaneous velocity at t = 2 s. Also find
c) the particle's acceleration function and
d) its instantaneous acceleration at t = 2 s.
Solution
First we note that the coefficients need units (these units are often omitted when calculating informally):
x
(
t
)
=
2
[
m
s
3
]
t
3
+
5
[
m
s
2
]
t
+
2
m
{\displaystyle x(t)=2\left[{\frac {\mathrm {m} }{\mathrm {s} ^{3}}}\right]\ t^{3}+5\left[{\frac {\mathrm {m} }{\mathrm {s} ^{2}}}\right]\ t+2\ \mathrm {m} \ }
a)
v
(
t
)
=
d
x
d
t
{\displaystyle v(t)=\,{\frac {dx}{dt}}}
=
d
d
t
(
2
[
m
s
3
]
t
3
+
5
[
m
s
]
t
+
2
m
)
{\displaystyle =\,{\frac {d}{dt}}\left(2\left[{\frac {\mathrm {m} }{\mathrm {s} ^{3}}}\right]\ t^{3}+5\left[{\frac {\mathrm {m} }{\mathrm {s} }}\right]\ t+2\ \mathrm {m} \ \right)}
=
|
6
[
m
s
3
]
t
2
+
5
[
m
s
]
|
¯
_
{\displaystyle ={\underline {\overline {\left|6\left[{\frac {\mathrm {m} }{\mathrm {s} ^{3}}}\right]\ t^{2}+5\left[{\frac {\mathrm {m} }{\mathrm {s} }}\right]\right|}}}}
b)
v
(
t
=
2
)
=
6
⋅
2
2
+
5
=
|
29
m
/
s
|
¯
_
{\displaystyle v(t=\mathrm {2} )=6\cdot 2^{2}+5={\underline {\overline {\left|29\ \mathrm {m} /\mathrm {s} \right|}}}}
c)
a
(
t
)
=
d
d
t
(
6
[
m
s
3
]
t
2
+
5
[
m
s
]
)
=
|
12
[
m
s
3
]
t
|
¯
_
{\displaystyle a(t)=\,{\frac {d}{dt}}\left(6\left[{\frac {\mathrm {m} }{\mathrm {s} ^{3}}}\right]\ t^{2}+5\left[{\frac {\mathrm {m} }{\mathrm {s} }}\right]\right)={\underline {\overline {\left|12\left[{\frac {\mathrm {m} }{\mathrm {s^{3}} }}\right]t\right|}}}}
d)
a
(
t
=
2
)
=
|
24
m
/
s
2
|
¯
_
{\displaystyle a(t=\mathrm {2} )={\underline {\overline {\left|24\ \mathrm {m} /\mathrm {s} ^{2}\right|}}}}
↑ https://en.wikibooks.org/w/index.php?title=Physics_with_Calculus/Mechanics/Motion_in_One_Dimension&oldid=2403866