Test 1 Review

Physics I Sp 01

 

Here is a sampling of problems that I would expect you to be able to solve, or answer.

 

1) Given 1 km = .62 miles, convert miles/hour to meters/min.

 

(miles/hr)(1 km/.62miles)(1000 m/1 km)(1 hr/60 min) = 26.9 m/min

 

2) When measuring a rectangle for a class project lazy Goofus only measures to the nearest cm and measures 4 x 5 for an area of 20 sq. cm.  Gallant takes his time

and measures to the nearest mm, getting 3.6 x 4.7, and he reports the area as 16.9 sq cm.

Which area measurement is more accurate, and why?  Who had points deducted for

having the wrong number of significant figures in his answer, and why?

 

Gallant’s answer is more accurate since the measurements were more precise,

however he should have given his answer to two significant figures, 17 sq cm.

 

3) If you read that a star is an order of magnitude of 5 larger than the earth, how many times the size of earth is the star?

 

Orders of magnitude are powers of ten.  The star is 100000 larger than earth.

 

4) Given A = 3i – 2j + 2k, and B = 2i + j - 2k Find:

 

a) 2A + B = (6 + 2)i + (-4 + 1)j + (4 – 2)k = 8i – 3j +2k

 

b)  = (3)(2) + (-2)(1) + (2)(-2) = 6 – 2 – 4 = 0

 

c)  = -6i + 10j +7k

 

d)   length of A is , the length of B is 3

 

e) The angle between A and B

 Since the dot product was 0, the vectors must be perpendicular, or 90 degrees.

 

5) A man walks 4 miles at 42 degrees N of E, then walks 7 miles 50 degrees W of N.  What is the magnitude and direction of his net displacement?

 

Let the first vector be A, and the second B.

 

Then, The resultant vector, C = A + B, has components

The magnitude of C is 7.6, and it is located 71.6 degrees N of E.

 

6) Describe the difference between translational motion and rotational motion. Why can

an object undergoing translational motion be treated as a “point” object?

 

Translational motion means the object moves with each of its parts undergoing

the same motion.  Since every part of the object moves the same, we can collapse

the object to one of its “points”.  Rotational motion differs since not all the parts

are moving the same.  Some may even be at rest while others rotate around them.

 

7) A car travels west for 50 miles at a speed of 70 miles/ hour, then immediately returns

at 60 miles per hour.  Find the total distance traveled and displacement of the car.  Find

the average speed, and average velocity of the car.

 

distance = 50 + 50 = 100 miles, displacement = 0 (returned to same spot)

 

average speed = 64.6 miles/hr (total time = 50/70 + 50/60 = 1.5476 hrs

   average velocity = 0

 

8) The position of a particle is given by the equation

 

a)Find the average velocity between t = 1, and t = 3.

 

Use the points (1, -3), and (3, 15) to find the slope = 18/2 = 9

 

b) Find the instantaneous velocity at t = .5

 

 used the derivative.

 

c) Find the instantaneous acceleration at t = 1

 

 used the second derivative

 

d) Find the average acceleration between t = 1, and t = 2.

 

Now use the first derivative for the velocity function to find the points (1, -1)

and (2, 8) and find the slope = 9/1 = 9

 

 

 

 

 

 

 

 

 

 

 

9) Consider the velocity (m/s) vs. time (s) graph.

 

 

 


v

 

10

 

5

 

   0               5            10             15                

 

 

a) Where is the acceleration positive, zero, negative? approximately:

Using interval notaion: positive (0, 3), zero at t = 3, negative (3, 18)

 

b) Does the particle ever reverse direction?  If so, at what time?

At t = 13

 

c) Estimate the displacement of the particle between 5 and 10 seconds.

This is the area under the graph from t = 5 to t = 10.   Break it into

a rectangle and another piece that looks like about half the rectangle

to get 25 + 25/2 = 75/2 sq units.

 

d) Estimate the average velocity between 0 and 5 seconds.

 

The acceleration is not constant so I cannot find this.  I would need the

displacement graph and then find the slope between those points.

 

e) Estimate the average acceleration between 0 and 5 seconds.

 

Use the points from (d) and find the slope = (10 – 3)/(5 – 3) = 7/2 m/s/s    

 

10) A car is initially moving at 112 km/h.  Find its acceleration and the time

taken to stop given that:

 

a) it brakes to a stop in 64m

initial velocity = 31.1 m/s,  final velocity = 0, and change in displacement is

64 m.  Use  to find a = -7.6 m/s/s, and use another

kinematics equation (your choice) to find t = 4.1 sec.

 

 b) it crashes into a wall and crumples in 1 m.

            a = -483.6 m/s/s   t = .06 sec  OUCH