**Questions are adapted from "College Physics: Open Stax College" (pg 81) and are used on WebAssign Under the title: "110(2.1) Position, Velocity, Acceleration, (Homework)"**

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Question

1.Find the following for path "D" in the figure

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Question

1.

a) the total distance traveled

b) the distance from start to finish

c) the displacement from start to finish

**Solution**

a) distance (scalar) from 9 to 3, then from 3 to 5.

(9 -3) + (5-3) =

**8(m)**

b) Only magnitude needs to be indicated. Direction doesn't matter.

distance from start to finish (straight line) = 9- 5=

**4(m)**

c) displacement is a vector (must indicate direction)

displacement = Δx = x

_{f}- x

_{i}= 5 - 9 =

**-4m**

**Question**

**2**. With the ability to measure the motion of land masses, continental drift has become an established fact. The North American and European continents are drifting apart at a rate of about 3cm/y. At this rate how long will it take them to drift 750 km further apart than at present?

______ yr

**Solution**

Uknown: "t"

Δx = 750 km (750,000 m)v = 3cm/y (0.03m/y)

NOTE: convert units to match (meter for distance, meters/year for velocity)

from equation: v = Δx/t

Solve for "t"

t = Δx/v

t = 750,000/0.03 (y) =

**25,000,000 (y)**

**Question**

**3**. A student drove to the university from her home and noted that the odometer reading of her car increased by 16.0 km. The trip took 17.0 min.

(a) What was her average speed?

_____ km/h

(b) If the straight-line distance from her home to the university is 10.3 km in a direction 25.0º south of east, what was her average velocity?

_____ km/h (25˚ S of E)

(c) If she returned home by the same path 7 h 30 min after she left, what were her average speed

_____ Km/h

(d) and velocity for the entire trip?

_____ km/h

**Solution**

a)

Known:

Distance: 16 km (odometer)

time: 17 min = 17/60 hours = 0.283333 hours

Known:

Distance: 16 km (odometer)

time: 17 min = 17/60 hours = 0.283333 hours

unknown: average speed:

speed = distance / time

speed = distance / time

speed = 16/0.28333 (km/h) =

**57.4712526 km/h**
b) velocity = displacement/time v=Δx/t

displacement is 10.3 km (in the direction of 25˚ S of E)

t = 0.28333 hours

displacement is 10.3 km (in the direction of 25˚ S of E)

t = 0.28333 hours

v = 10.3/0.28333 =

**36.35336 km/h**
c)

d = 16 X 2 = 32km

t = 7.5 h

v = d/t = 32/7.5 =

**4.2667 km/h**
d)

v=Δx/tΔx = x

v=Δx/tΔx = x

_{f}- x_{i = 0 - 0 = 0 km (she started and ended at the same position)}_{v = 0/7.5 = 0 km}

_{}

_{Question}

**4.**A commuter backs her car out of her garage with an acceleration of 1.50 m/s2.

(a) How long does it take her to reach a speed of 2.4 m/s?

_____ s

(b) If she then brakes to a stop in 0.800 s, what is her deceleration?

_____ m/s

^{2}

^{Solution}a)

v

_{f}= 2.4m/s

vi = 0 m/s

a = 1.5 m/s

^{2}

a = Δv/t = (v

_{f}- v

_{i})/t

Solve for "t"

t = (v

_{f}- v

_{i})/a = (2.4 - 0) /1.5 =

**1.6 seconds**

b)

vf = 0

vi = 2.4 m/s

t = 0.8 s

a = Δv/t = (v

_{f}- v

_{i})/t = (0 - 2.4) /0.8 =

**-3 m/s**

^{2}

^{Question}^{5. }Assume that an MX missile goes from rest to a suborbital speed of 9.50 km/s in 30.0 s (the actual speed and time are classified).

a) What is its average acceleration in m/s

^{2}

_____

b) and in multiples of g?

_____

**Soultion**a)

v

_{f}= 9.5km/s = 9,500 m/s

vi = 0 m/s

t = 30 sec

a = Δv/t = (v

_{f}- v

_{i})/t = (9500 - 0)/30 =

**316.67 m/s**

^{2}b) g = 9.8 m/s

^{2}

in multiples of "g"

a/g = 316.67 / 9.8 =

**32.3129 g**

A light-rail commuter train accelerates at a rate of 1.55 m/s2. How long does it take it to reach its top speed of 80.0 km/h starting from rest?

ReplyDeletes

(b) The same train ordinarily decelerates at a rate of 1.65 m/s2. How long does it take to come to a stop from its top speed?

s

(c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s2?

m/s2

anonymous:

DeletePART A:

a = (vf - vi)/t (change of time) (acceleration = change of velocity over change of time)

vf = 80 km/h

vi = 0 km/h

a = 1.55 m/s^2

1. Convert 80km/h into m/s

km/h X 1/3600 X 1000/1

in other words, 80/36 = 22.22m/s

2. Solve the equation for "t"

t = (vf - vi)/a

3. Plug in the vf and vi (in m/s) and plug in a -- The answer will magically appear

t = (22.22 - 0) / 1.55 = 14.34 seconds

part B: Same thing as before, except divide 22.22 by 1.65

t = 13.47s

Part C

22.22/8.3 = 2.68m/s^2