CSULA PHYS211 Summer’12 Chapter 2: Motionin One Dimension Homework Assignment #1 ConceptualProblems 1. Driver A is cruising along enjoying the fall colors. Driver Bstarts her car at the instant he...

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CSULA PHYS 211 Summer’12


Chapter 2: Motion in One Dimension Homework Assignment #1


Conceptual Problems



1.
Driver A is cruising along enjoying the fall colors. Driver B starts her car at the instant he passes her. Their velocities are shown as functions of time in the graph below. At what instants in time on the graph are drivers A and B side by side?



a.0 s, 2 s

b.
0 s, 4 s
c.
2 s, 4 s




d.
2 s, 6 s

e.
4 s, 6 s




2.

A particle moving along the
xaxis has a position given by
x
= (24t
– 2.0t
3) m, where
t
is measured in s. What is the



magnitude of the acceleration of the particle at the instant when its velocity is zero?




a.

24 m/s2b.0
c.12 m/s2d.48 m/s2
e.36 m/s2





Open-Ended Problems




3.
Starting from rest, a car travels 1,350 meters in 1.00 minute. It accelerated at 1.0m/s2 until it reached its



cruising speed. Then it drove the remaining distance at constant velocity. What was its cruising speed?



4.A car originally traveling at 30m/s manages to brake for 5.0 seconds while traveling 125 m downhill. At that point the brakes fail. After an additional 5.0 seconds it travels an additional 150 m down the hill. What was the acceleration of the car after the brakes failed?



5.
A speedy tortoise can run with a velocity of 10 cm/s and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 2.0 minutes. The tortoise wins by a shell (20 cm). What was the length of the race?





Answered Same DayDec 22, 2021

Answer To: CSULA PHYS211 Summer’12 Chapter 2: Motionin One Dimension Homework Assignment #1...

David answered on Dec 22 2021
119 Votes
1. The graphs are not given.
2. We have x=24t-2t3
Velocity v=dx/dt=24-6t2
Velocity is zero at
t=2 sec.
Acceleration a=dv/dt=-12t
Acceleration is -24 m/sec2 when the velocity is zero.
Answer=(a).
3. let x = time of acceleration and y = time at constant speed
cruising speed, v is
v = u + at = 0 + 1*x = x
distance covered during the acceleration phase, sa, is
sa = ut + 0.5at^2 = 0 + 0.5*1*x^2 = 0.5x^2
distance covered at the cruising speed, sc is
sc = vy = xy
x + y = 60 (remember to convert minutes to seconds)
sa + sc = 1350
so we have
0.5x^2 + xy = 1350
and y = 60-x
that is...
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