WORKSHEET 2: 1-Dimensional Motion
PROBLEM SOLVING TACTICS: Tactic 3 Derivatives and Slopes
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Each derivative can be interpreted as the slope of a curve at that point.
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The plot below shows x(t) for a moving particle. To find the velocity at
t = 1 s, locate the x value corresponding to t = 1 s and draw a line tangent
to that point on the curve (see figure).
-
Using this tangent line, draw a convenient right triangle with sides parallel
to the axes and hypotenuse at the tangent line.
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The slope = Dx/Dt
= (5.5 m 2.3 m)/(1.8 s 0.3 s) = 3.2 m/1.5 s = +2.1 m/s.
Problem 1: Spotting a police car, you brake your Porsche from
75 km/h to 45 km/h over a displacement of 88 m.
(a) What is your acceleration, assuming that it is constant?
SOLUTION:
Assuming constant acceleration, we can use the appropriate equation from
the table above.
75 km/h = 20.8 m/s and 45 km/h = 12.5 m/s
(b) What is the elapsed time?
SOLUTION:
v = v0 +at
t = (v v0 )/a = (12.5 m/s 20.8 m/s)/(-1.6 m/s2
)
t = -5.2 s
(c) If you continue to slow down with the acceleration calculated in
part (a), how long will it take to bring your car to a stop from 75 km/h?
SOLUTION:
This time v0 = 75 km/h = 21 m/s and v = 0 and the acceleration
is still a = -1.6 m/s2 .
t = (v v0 )/a = (0 m/s 20.8 m/s)/(-1.6 m/s2 )
t = 13 s.
(d) In (c) what is the total distance covered?
SOLUTION:
Several equations can be used for this question, lets use the following:
x x0 = ½ (v + v0 )t
x x0 = ½ (21 m/s + 0)(13 s) = 136.5 m = 1.4 x 102
m.
Problem 2: The minimum distance required to stop a car
moving at 35 mi/h is 40 ft. What is the minimum stopping distance
for the same car moving at 70 mi/h assuming the same rate of acceleration?
Solution:
Step 1: Find the acceleration of the car
using the initial information: x - x0 = 40 ft, vf =
0 and vi = 35 mi/h
vf2 = vi2 -2a(x - x0)
a = (vf2 - vi2 )/2(x - x0)=[(0
- (35mi/h)2)/80ft][(5280ft/1mi)(1h/3600 s)2]
a = -32.9 ft/s2
Step 2: Using this acceleration, find (x
- x0) for the car traveling at 70 mi/h
(x - x0) = (vf2 - vi2
)/2a
= [0 -(70 mi/h)2/2(-32.9 ft/s2)][(5280ft/mi)(1h/3600
s)2]
= 160 ft
Problem 3: A "superball" is dropped
from a height of 2.00 m above the ground. On the first bounce, the
ball reaches a height of 1.85 m, where it is caught.
a) Find the velocity of the ball
just as it makes contact with the ground
Solution:
Knowns: vo = 0, x - xo = -2.00 m
v2 = vo2 +2a(x - xo) = 0 +2(-9.80
m/s2)(-2.00 m)
v = 6.26 m/s (downward)
b) Find the velocity of the ball
on the bounce
Solution:
Knowns: v = 0, x - xo = 1.85 m
v2 = vo2 +2a(x
- xo) = 0 +2(-9.80 m/s2)(1.85 m)
vo = 6.02 m/s (upward)
c) Total time of flight neglecting
time of contact with the ground.
Solution:
t1 = (Dv1)/a
= (6.26 m/s)/(9.80 m/s2) = 0.64 s
t2 = (Dv2)/a
= (6.02 m/s)/(9.80 m/s2) = 0.61 s
t = t1 + t2 = 1.25 s
Problem 4: A worker drops a wrench down the elevator shaft of
a tall building.
(a) Where is the wrench 1.5 s later?
SOLUTION:
Knowns: v0 = 0, y0 = 0, t = 1.5 s
y = y0 + v0 t ½ gt2 = 0 + 0
½ (9.80 m/s2 )(1.5 s)2 = 11 m
(b) How fast is the wrench falling just then? (after t = 1.5 s)
SOLUTION:
v = v0 gt = 0 (9.80 m/s2)(1.50 s) = -14.7 m/s
Problem 5: A pitcher tosses a baseball straight up, with an
initial speed of 12 m/s.
(a) How long does the ball take to reach its highest point?
SOLUTION:
Knowns: v = 0 at the highest point, v0 = 12 m/s
v = v0 gt
t = (v0 v)/g = (12 m/s 0)/(9.80 m/s2 ) = 1.2
s
(b) How high does the ball rise above its release point?
SOLUTION:
y = y0 + v0 t ½ gt2 = 0 + 12
m/s(1.2 s) ½ (9.80 m/s2 )(1.2 s)2 = 14.4
m 7.1 m = 7.3 m
(c) How long will it take for the ball to reach a point 5.0-m above
its release point?
SOLUTION:
Knowns: y y0 = 5.0 m, v0 = 12 m/s
y = y0 + v0 t ½ gt2
5.0 m = (12 m/s)t - (4.90 m/s2 )t2 , the only way
to solve this is using the quadratic equation.
4.90 t2 12t + 5.0 = 0
