Displacement, velocity, and acceleration problem solutions
This paper is a series of five physics problems. The problems cover vectors, deceleration, displacement, and acceleration. The paper includes a vector diagram to explain a Cartesian vector problem. The paper shows all of the work for the problems in the paper and puts the answers in bold.
✍️ How to write this paper — guide, tools & examples ▾
Displacement, Velocity, And Acceleration
Work out the following problems. Be sure to show your work in detail. See the uploaded files for examples of how the problems should be solved and presented.
A car is driving along a straight section of highway between Oklahoma City and Tulsa. At one point in time, the car is 50 miles from Tulsa. At a later point in time, it's 40 miles from Tulsa. What is its displacement?
s = x-x0
s=10 miles
An aircraft flies 20 km on a course of 030 degrees, then 30 km on a course of 270 degrees. What is its distance and bearing from its takeoff point? (Hints: Make a sketch. Convert flight legs to Cartesian vectors in standard position. See the "Overview of Vectors" handout.)
km at 30O
km at 270O
VR (XRYR)= V1 = V2
Because the second vector is at 270O it makes a right angle with the first vector, so that the length of the resulting vector, VR can be determined with the simple equation a2 + b2= c2
202 + 302 = c2
400+900= c2
c= 36.06km
Then, one can determine the angle of a, the angle including the 30O of the initial angle and whatever the angle is below the x-axis by looking at the sine, which is the opposite leg (30km) over the hypotenuse (36km).
sin O. Of a= 30 / 36.06 = 8.32, therefore a= 56.30 O
56.30 -- 30 = 26.30 O
Therefore, the vector is traveling at 26.30 O. clockwise from the x-axis, but coordinates are measured counter-clockwise in the Cartesian plane, so 360 O -- 26.30 O = 333.77 O
Therefore, the resulting vector is 36.06 km at 333.77 O
3. A snail travels for a year on a straight line, covering 2.7 km. What is its average velocity, in m/s?
2.7km/yr * 1000m/1km * 1 year/365 days * 1 day / 24 hours * 1 hour / 3600 sec=
.0000856164 m/sec
8.56 x 10-5 m/s
4. The parachute on a top fuel dragster slows the car from 300 mph to 100 mph in 5 seconds. What is the average deceleration, in gs (acceleration of gravity)? (Hint: Use an online coversion app, and work in MKS [metric] units.0
d= v1-v2 / t d= 300 mph- 100 mph / 5 sec
d= 200 mph / sec convert:
200 mph* 1m/s = 89.41 m/s
2.237 mph
89.41m/s / 5s = 17.88 m/s2
Create your account
Always verify citation format against your institution’s current style guide requirements.