PHYS101
Motion in One Dimension
Spring 2014
Motion in One Dimension
1. The position of an object moving along an x axis is given by
x (t) = 3t − 4t2 + t3 ,
where x is in meters and t in seconds. Find the position of the object at the
following values of t:
(a) 1 s,
Solution:
Evaluation of x (t) at t = 1s yields x (1s) = 3 − 4 + 1 = 0.
(b) 2 s,
Solution:
Evaluation of x (t) at t = 2s we get x (2s) = 3(2) − 4(2)2 + (2)3 = −2m.
(c) 3 s, and
Solution:
Evaluation of x (t) at t = 3s we have x (3s) = 0.
(d) 4 s.
Solution:
Evaluation of x (t) at t = 4s gives x (4s) = 12m.
(e) What is the object’s displacement between t = 0 and t = 4s?
Solution:
In order to find the displacement in the time between t = 0 and t = 4s
we calculate the difference of the positions at these times as ∆x = x (4s) −
x (0) = 12m.
(f) What is its average velocity for the time interval from t = 2s to t = 4s?
Solution:
First we need to calculate the displacement in the time interval between
t = 2s and t = 4s by subtracting the position at t = 2s from the position at
t = 4s.
∆x = x (4s) − x (2s) = 12m − (−2m) = 14m.
Then we calculate the average velocity by
v avg =
∆x
14m
=
= 7m/s.
∆t
2s
(g) Graph x versus t for 0 ≤ t ≤ 4s and indicate how the answer for 1f can be
found on the graph.
Solution:
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Department of Physics, Eastern Mediterranean University
Page 1 of 12
PHYS101
Motion in One Dimension
Spring 2014
The position of the object for the internal 0 ≤ t ≤ 4 is plotted below. The
slope of the straight line drawn from the point at (t, x ) = (2s, −2m)to (4s, 12m)
represents the average velocity calculated in part 1f.
2. You are to drive to an interview in another town, at a distance of 300 km on an
expressway. The interview is at 11:15 A.M. You plan to drive at 100 km/h, so
you leave at 8:00 A.M. to allow some extra time. You drive at that speed for the
first 100 km, but then construction work forces you to slow to 40 km/h for 40
km. What would be the least speed needed for the rest of the trip to arrive in
time for the interview?
Solution: The first part of the trip (at 100 km/h) takes 1 hour, and the second
part (at 40 km/h) also takes 1 hour. Expressed in decimal form, the time left is
1.25 hour, and the distance that remains is 160 km. Thus, an average speed of
v = (160km)/(1.25h) = 128km/h is necessary to be on time at the interview.
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2014
Department of Physics, Eastern Mediterranean University
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PHYS101
Motion in One Dimension
Spring 2014
3. The position function x (t) of a particle moving along an x axis is x = 4.0 − 6.0t2 ,
with x in meters and t in seconds.
(a) At what time and where does the particle (momentarily) stop?
(b) At what negative time and positive time does the particle pass through the
origin?
(c) Graph x versus t for the range −5s to +5s.
(d) To shift the curve rightward on the graph, should we include the term −20t
or the term 20t in x (t)?
(e) Does that inclusion increase or decrease the value of x at which the particle
momentarily stops?
Solution:
We use the functional notation x (t), v(t), and a(t) in this solution, where the
latter two quantities are obtained by differentiation:
v(t) =
dx (t)
= −12t
dt
and
a(t) =
(unit of v(t) is
dv(t)
= −12
dt
m/s
m
therefore − 12 t)
,
s
(unit of a(t) is
m
)
s2
(a) From v(t = 0) = 0 we find it is (momentarily) at rest at t = 0.
(b) For the time at which the particle passes through the origin we have to
calculate:
√
√
x (t) = 4 − 6t2 = (2 + 6t)(2 − 6t) = 0
A product is equal to zero if one of the factors is zero, therefore we get
2
t1/2 = ± √ = ±0.82s
6
(c) The graph of x (t) for t ∈ [−4s, 4s] is
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PHYS101
Motion in One Dimension
Spring 2014
x(m)
-4
0
-20
4
t(s)
-40
-60
-80
-100
(d) In order to decide whether adding +20t or −20t shifts the graph to the
right, let us consider the general rule as we have learned it from PreCalculus. Shifting the graph to the right by a units, with a > 0 is performed by
the substitution t → t − a. Applying this to our problem we get
x (t − a) = 4 − 6(t − a)2 = 4 − 6t2 + 12ta − 6a2 .
Therefore obviously we have to add +20t to complete this operation. Furthermore we can see that by adding 20t corresponds to shifting the graph
by 5/3 units to the right.
(e) Shifting the graph to the right by adding +20t, the maximum, which denotes the point where the velocity becomes momentarily zero shifts also by
5/3 units to the right. But the x value, where v(t) becomes zero does not
change. as we can see from the figure below.
x(m)
x(m)
-4
0
-20
4
t(s)
-4
0
-20
-40
-40
-60
-60
-80
-80
-100
-100
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4
t(s)
Page 4 of 12
PHYS101
Motion in One Dimension
Spring 2014
4. The position of a particle moving along the x axis is given in centimeters by
x = 9.75 + 1.50t3 , where t is in seconds.
Calculate
(a) the average velocity during the time interval t = 2.00s to t = 3.00s;
(b) the instantaneous velocity at t = 2.00s; the instantaneous velocity at t =
3.00s; the instantaneous velocity at t = 2.50s; and
(c) the instantaneous velocity when the particle is midway between its positions at t = 2.00s and t = 3.00s.
(d) Graph x versus t and indicate your answers graphically.
Solution:
(a) The average velocity during the time interval 2.00 ≤ t ≤ 3.00s is
v avg =
x (3.00s) − x (2.00s)
50.25cm − 21.75cm
∆x
=
=
= 28.5cm/s
∆t
3.00s − 2.00s
3.00s − 2.00s
(b) The instantaneous velocity for all t is
v(t) =
dx (t)
= 4.5t2 .
dt
In order to determine v(t) at t = 2.00s,t = 3.00s, and t = 2.50s we just have
to evaluate v(t) at these values and get :
v(2.00s) = 4.5(2.00s)2 = 18.0cm/s
v(3.00s) = 4.5(3.00s)2 = 40.5cm/s
v(2.50s) = 4.5(2.50s)2 = 28.1cm/s
(c) In order to calculate the instantaneous velocity on the midway between
x (2.00s) and x (3.00s) we first have to calculate xm as following:
xm =
21.75cm + 50.25cm
x (2.00s) + x (3.00s)
=
= 36cm
2
2
Now we have to calculate the time tm , where x (tm ) = 36cm.
r
3 36 − 9.75
xm = 9.75 + 1.5t3m = 36cm ⇐⇒ tm =
= 2.596s
1.5
Then we get for the velocity v(tm )
v(tm ) = 4.5(2.596s)2 = 30.3cm/s.
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Department of Physics, Eastern Mediterranean University
Page 5 of 12
PHYS101
Motion in One Dimension
Spring 2014
(d) The slope of the black line indicates the average velocity of the particle
calculated in (a), the slope of the blue line corresponds to v(2.00s), the slope
of the green line corresponds to v(2.50s), and the the slope of the red line
corresponds to v(3.00s) as calculated in part (b).
x(m)
50
40
30
20
10
0
1
2
3
4
t(s)
5. The position of a particle is given by x = 20t − 5t3 , where x is in meters and t is
in seconds.
(a) when, if ever, is the particle’s velocity zero?
(b) When is its acceleration zero?
(c) For what time range (positive or negative) is the acceleration a(t) negative
or positive?
(d) Graph x (t), v(t), and a(t).
Solution:
We use the functional notation x(t), v(t), and a(t) in this solution, where the latter
two quantities are obtained by differentiation:
v(t) =
dx (t)
= −15t2 + 20
dt
and
a(t) =
dv(t)
= −30t
dt
(a) From 0 = −15t2 + 20, we see that the only
√ positive value of t for which the
particle is (momentarily) stopped is t = 20/15 = 1.2 s.
(b) From 0 = 30t, we find a(0) = 0 (that is, it vanishes at t = 0).
(c) It is clear that a(t) = −30 t is negative for t > 0. The acceleration a(t) =
−30 t is positive for t < 0.
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Department of Physics, Eastern Mediterranean University
Page 6 of 12
PHYS101
Motion in One Dimension
Spring 2014
(d) The graphs are shown below.
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Page 7 of 12
PHYS101
Motion in One Dimension
Spring 2014
6. The brakes on your car can slow you at a rate of 5.2m/s2 .
(a) If you are going 137km/h and suddenly see a state trooper, what is the
minimum time in which you can get your car under the 90 km/h speed
limit? (The answer reveals the futility of braking to keep your high speed
from being detected with a radar or laser gun.)
(b) Graph x versus t and v versus t for such a slowing.
Solution:
(a) Substituting v0 = 137km/h = 38.1m/s , v = 90km/h = 25m/s and a =
−5.2m/s2 into v = v0 + at, we obtain
t=
25m/s − 38m/s
= 2.5s
−5.2m/s2
We take the car to be at x = 0 when the brakes are applied (at time t =
0). Thus, the coordinate of the car as a function of time is given by x =
(38m/s)t + 21 (−5.2m/s2 )t2
(b) Graph of x (t) and v(t)
x(m)
100
v(m/s)
50
80
40
60
30
40
20
20
10
0
1
2
3
4
5
t(s)
0
1
2
3
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4
5
t(s)
Page 8 of 12
PHYS101
Motion in One Dimension
Spring 2014
7. A car moves along an x axis through a distance of 900 m, starting at rest (at
x = 0) and ending at rest (at x = 900m). Through the first 14 of that distance, its
acceleration is +2.25m/s2 . Through the rest of that distance, its acceleration is
−0.750m/s2 . What are
(a) its travel time through the 900 m and
(b) its maximum speed?
(c) Graph position x, velocity v, and acceleration a versus time t for the trip.
Solution:
(a) First we have to find the time for the first 1/4 of the trip’s distance ∆x1 =
900/4m = 225m. The acceleration in this part of the trip is a1 = 2.25m/s2
starting from rest, i.e. v01 = 0. Then we get
1
∆x1 = a1 t1 2 ⇐⇒ t1 = ±
2
s
2∆x1
=
a1
r
√
2 · 225m
=
10
2s = 14.14s
2.25m/s2
In the second part the car slows down over a distance ∆x2 = 900m −
225m = 675m with a2 = −0.750m/s2 . But before we can calculate the
time for this part we have to determine the velocity at x = 225m in order to
be able to calculate the time.
√
v02 = v1 (t1 ) = a1 t1 = 2.25m/s2 · 10 2s = 31.82m/s
Now we can calculate the time for the second part of the trip
1
∆x2 = v02 t2 + a2 t22
2
v0
t2 = − 2 +
a
s
v202 + 2a2 ∆x2
a22
= 42.43s.
So the total travelling time is
t = t1 + t2 = 14.14s + 42.43s = 57.57s
(b) The maximum speed is at t = 14.14s, i.e. when the car has travelled 225m
and is: v02 = 31.82m/s.
(c) Graphics for x (t), v(t), and a(t).
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Department of Physics, Eastern Mediterranean University
Page 9 of 12
PHYS101
Motion in One Dimension
x(m)
Spring 2014
a(m/s2)
v(m/s)
30
800
2.0
25
600
1.5
20
400
1.0
15
0.5
10
200
10
20
30
40
50
t(s)
5
10
20
30
40
50
t(s)
0.5
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Department of Physics, Eastern Mediterranean University
10
20
30
40
50
Page 10 of 12
t(s)
PHYS101
Motion in One Dimension
Spring 2014
8. A gangster throws a stone vertically downward with an initial speed of 12.0m/s
from the roof of a building, 30.0m above the ground.
(a) How long does it take the stone to reach the ground?
(b) What is the speed of the stone at impact?
Solution:
Before we start solving this problem, we need define the frame of reference.
y
0
(a) In order to calculate the stone’s flight time until the stone reaches the ground,
i.e. y(t) = 0 we have to solve
1
y(t) = y0 + v0 t + at2
2
for t.
t=−
v0
±
a
s
v20 − 2a(y0 − y(t))
a2
with v0 = −12m/s , y0 = 30m, y(t) = 0 and g = −9.8m/s2 we get t =
1.53s.
(b) The speed of the stone at impact is v(t = 1.53s) = v0 + at where v0 =
−12m/s , g = −9.8m/s2 and t = 1.53s. So easily found that v(t = 1.53s) =
−27.1m/s.
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Department of Physics, Eastern Mediterranean University
Page 11 of 12
PHYS101
Motion in One Dimension
Spring 2014
9. A rock is thrown vertically upward from ground level at time t = 0. At t = 1.5s
it passes the top of a tall tower, and 1.0s later it reaches its maximum height.
What is the height of the tower?
Solution:
Again we have to define the frame of reference before performing the calculations.
y
0
The rock reaches its highest point after t = 2.5s, where the velocity of the rock is
v(2.5s) = 0. So we can calculate the initial velocity as
v(t) = v0 + at =⇒ v0 = v(t) − at = 0 − (−9.8
m
m
)2.5s = 24.5 .
2
s
s
Now we can calculate the height of the tower as the rock passes the top of the
tower after t = 1.5s. So we get for the height of the tower.
m
1 m
1
y(t) = y0 + v0 t + gt2 = 0 + 24.5 1.5s − 9.8 2 (1.5s)2 = 25.7m.
2
s
2 s
So the height of the tower is 25.7 m.
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2014
Department of Physics, Eastern Mediterranean University
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