1st year college physics questions
Transcript of 1st year college physics questions
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velocity (m/s)
Time (s)
5
10
15
5 10 15
15 20 25 30
-5
PROBLEM 2.1. ANSWER
THE FOLLOWING
QUESTIONS ABOUT A
PARTICLE WHOSEVELOCITY IS PLOTTED
VERSUS TIME TO LEFT
(ORANGE LINE)
(a) Use the velocity graph above to plot quantitatively a graph of
acceleration versus t ime, with correct scales on both axes.
acceleration (ms-2)
Time (s)
2
5 10 15
15 20 25 30
-2
2 1
2 1
5 5Motion from t=0 to t=10s: accln is 0 ( constant)10 0
v va t t
= = = =
215 5Motion from t=10s to t=15s: accln is 2 / ( constant)15 10
f i
f i
v va m s
t t
= = = =
2( 5) 15 20Motion from t=15s to t=25s: accln is 2 / ( constant )25 15 10
f i
f i
v va m s
t t
= = = = =
( ) 2( 5) 5 0Motion from t=25s to t=30s: accln is 0 / ( constant )30 25 5
f i
f i
v va m st t
= = = = =
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(b) Find the displacement (change of position) between t = 10 sec and t = 15 sec.
25
velocity (m/s)
Time (s)
5
10
15
5 10 15
15 20 30
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WORKING TO GET
AREA BETWEEN
VELOCITY VS T
GRAPH AND TIME
AXIS
Displacement f ix x
area between v t graph and t axis
= =
+
= (base)(ht) +(1/2)(base)(ht)
= 5X5 + (1/2)(5)(10)= 50 m
(c) Find the displacement between t = 20 sec and t = 25 sec. +
= (1/2)(base)(ht) +(1/2)(base)(ht)
= (1/2)(2.5)(5) + (1/2)(2.5)(-5)= 6.25 6.25 = 0 m
Note area below axis is negative
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Answer 2. The shortest possible runway has a length equal to the distance s
travelled by the plane while coming to rest. (But this would leave the pilot withno room for error the wheels would have to touch down exactly at the near end
of the runway!)
This is a problem of constant acceleration so we use one of the Eqs from slide #
28 of the Mechanics A Lecture Slides.
Given: Initial velocity v0 = 220 kph = 220/3.6 m/s = 61.11 m/s (converted to SIunits)
Time t = 29 sec
Final velocity = 0 (because the plane comes to rest)
Wanted: displacement (distance) s = ?
The acceleration a is missing neither given nor wanted.
Therefore we use Eq 3 from Lecture Slide 29, , which also has a missing:
886 metre.
This is the shortest possible length of a runway
Since only 2 signi ficant figures were given for the time, best quote 8.9X102 m
Problem 2
A jetliner touches down at 220 km/h, reverses its engines to provide braking, andcomes to a halt 29 sec later. What is the shortest runway on which this aircraft
can land, assuming constant deceleration starting from touchdown?
0 61.11 0 292 2
v vs t
+ += = =
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Problem 3
Launched from the ground, a rocket accelerates vertically upwards at 4.6 ms-2. It
passes through a band of clouds 5.3 km thick, extending upwards from an altitude
of 1.9 km. How long is it in the clouds?
y=0
y=1.9 km
y=7.2 km
In this problem we are dealing with data involving three
spatial locations A(y=0), B(y=1.9 km) and C(y = 1.9+5.3 km = 7.2 km). A single equation-of-constant-
acceleration can only deal with two spatial locations (an
initial point and a final point). Therefore we will need TWO
applications of constant-acceleration equation(s).
There are (at least) two different strategies that will solve
this problem.Strategy 1.
(a) Apply eq of const accel to find time tAB from ground to
cloud bottom.
(b) Apply eq of const accel to find time tAC from ground to
cloud top.
(c) Time in cloud is tBC= tAC - tABStrategy 2.
(a) Solve motion from A to B to get velocity at B, vB.
(b) Use vB as the initial velocity in motion B to C, hence
getting tBC.
A
B
C
groun
dd
cloud
Solution 3
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Solution of Problem 3 via Strategy 1
(a) For const.-acceln. motion A to B (choosing upwards as positive):v0 = 0, a = +4.6 ms
-2, s = 1.9 km = 1.9103 m. t = ?
Final velocity v is missing so use s = v0t + at2
1.9103 = 0 + (4.6)t2
t2 = (1.9103)(2)/(4.6) = 826.09
t = (826.09) = 28.742 sec = tAB(b) For const.-acceln. motion A to C (choosing upwards as positive):
v0 = 0, a = +4.6 ms-2, s = 7.2 km = 7.2103 m. t = ?
By same working as above tAC = [(7.2103)(2)/(4.6)] = 55.95 sec.
(c) Then time in cloud = tAC tAB = 55.95 28.742 = 27.208 sec
Solution of Problem 3 via Strategy 2.(a) for const acel motion from A to B (upwards positive) : v0 = 0, a = +4.6 ms
-2, s = +1.9103 m; v = ?
Here t is missing, so we use v2 = v02 +2as
= 02 + 2(4.6)1.9103 = 17480
v = 17480 = 132.211
(b) for const acel motion from B to C (upwards positive): v0 = 132.211, a= +4.6 ms-2, s = +5.3103 m, t = ?
Here v is missing so we use s = v0t + at2(a)t2 + (v0)t + (s ) = 0
Solution of quadratic for t f rom quadratic formula:
= 27.208 sec ( + solution used: - solution unphysical)
Ans: by either Strategy, solut ion is Rocket Spends 27 sec in the cloud layer(Quot ing more than 2 decimal places is not just ified by precision of input data)