Need help with physics question: A car travels 35 km west, 90 km north. What distance did it travel?
I know these questions are very simple and this is just review for me, but Im forgetting the concept, of what I should be doing with the numbers, and what “direction” the final answer can be, and how to determine this. Should I be doing it like this: d=d1 – d2, d=35-90, d= -55 (west or east??) or is this even right? Other questions that I have trouble with are: An airplane flies 250 km at 300 m/s. How long does this take? A skydiver falls 3 km in 15s. How fast are they going? (I keep forgetting the formula to get speed given distance and time or speed and time or time and distance). A car travels 35 km west, 90 km north in two hours. What is its average speed? A car travels 35 km west, 90 km north in two hours. What is its average velocity?
Ahh sorry I ment to say displacement for the first one, not distance. Corrected: A car travels 35 km west, 90 km north. What is it’s displacement? What would be a detailed process to get the displacement of this question and similar questions like this.
sharethis
the resultant formula is (square root)p^2+q^2+2pqcos@therefore the answer to above question is (sqr root) of 35^2+90^2 =96. 5kms in northwest.
Two things to keep in mind: velocity is a vector, i. e. , it has a direction associated with it. and then velocity = distance/ time or time = distance/velocity or distance = velocity * time. When solving the first question you have take direction into account. The car goes 35 km west and then 90 km north so the final direction was actually northwest. Speed is just the total distance in the total time = 125 km/2 hours = 62. 5 kmph. However velocity is with the final direction and reflects the straightline distance. This can be calculated by looking at the geometry of the path traveled – it is essentially a right triangle so the straightline distance travelled is sqrt( 35^2 + 90^2) = 96. 57 km so the velocity is 96. 57 km/ 2 hours = 43. 29 kmph in the northwest direction.
for your first question, no. . . . it should be like this:
use pythagorean theorem to find the hypotenuse(distance travelled):
d=35+90
d=125km
for airplane question. . .
we assume that there is no acceleration. . . :
250km=250000m
d=Vi*t+. 5*a*t^2
250000=300*t+0
t=250000/300
t=833. 33s
for skydiver question:
d=Vi*t-. 5*g*t^2
3000=Vi*15-. 5*9. 8*15^2
Vi=4102. 5/15
Vi=273. 5m/s
for the last question:
if we assume that they both have the same time:
d=Vi*t
(35000+90000)meters=Vi*120s
Vi=1041. 67m/s
**average speed and average velocity is the same. . .
The first part is simple math.
A right angled triangle and you use pythagorus to find the net distance.
For the other ones convert to standard units ie km and m/s
go to km/hr or m and remember the simple distance = speed X time. Then you switch the formula around for the speed or time.
For your first question
Problems like these usually involve vectors, which are physical quantities of magnitude in which the direction is specified (e. g a velocity of 50m/s due north) as opposed to scalars which do have magnitude but no specified direction (e. g a speed of 50m/s)
Now distance is a scalar and its vector counterpart is displacement.
So if you are looking for total distance travelled you would simply add the separate distances, but to find the resultant displacement you would need to add the vetors using a special parallelogram law.
So in this case the DISTANCE would be found by the formula
d = d1 + d2 = 35km + 90km = 125km
But the DISPLACEMENT can be found by pythagoras’ theorem as the vector quantities are at right angles forming a right angled triangle with the hypotenus being the resultant displacement you wish to find.
so d = sqrt(d1^2 + d2^2) = sqrt(1225 + 8100) = 96. 6km and the direction of such a vector would be given as an angle to a particular body of reference, usually the horizontal, but in this case it would be ‘at 45 degrees to the horizontal’.
For your second question
The formula connecting distance (d), speed, (s) and time (t)
is: d = s x t
If you have any of the two variables you can rearrange the equation to find the third. But remember, to take care with the units, you must first choose whether to work in kilometers, meters for length and seconds, minutes, hours etc. for time intervals, for simplicity, it is easier to convert all distances to meters, all time intervals to seconds and all speeds to meters per second and work from there.
So for example
An airplane flies 250 km at 300 m/s. How long does this take?
Converting the 250km to meters for simplicity gives 250000m
Now using the formula t = d/s
t = 250000m/300m/s
t = 833. 3s
For your third question
The difference between speed and velocity is that the latter is a vector quantity which I have explained above, but the equations for speed and velocity are pretty much the same, except speed is distance/time and velocity is displacement/time
To find average speed you would then simply add the distances (explained above) and divide by time
and for average velocity you would find the displacement (by pythagoras’ theorem) and divide by time.
Try to imagine and draw this travel in the form of a triangle. First a line to the left (which represents 35km) and from there a line up (which represents 90km). If you now connect your end point to your starting point you get a triangle, right? The distance between your starting point and your end point (this is the distance travelled) is exactly the line you’ve just drawn to get the triangle.
This triangle has an angle of 90 degrees (i. e. right angle)
For these kinds of triangles you use the Pythagoras theorem, which says that (horizontal line)^2 + (vertical line)^2 = (not-straight line)^2
Please note that “not-straight line” represents the distance travelled in your triangle.
Thus: (35km)^2 + (90km)^2 = (distance)^2
distance = SQRT (35^2 + 90^2) = 96. 6 km
The formula for speed (or velocity, is the same thing) = distance travelled / time
Thus speed = distance / time
distance = speed * time
time = distance / speed
AIRPLANE
So, let’s start with the airplane flying 250km at 300 m/s.
For this one, you can use a ‘trick’, which goes as follows:
you travel 300 meters in 1 second
i. e. you travel 250km in x seconds
You have to make sure that the same ‘kind of thing’ is on the same side of your 2 expressions:
300 m in 1 sec
250 km in x sec
(x sec since you want to know how long it took)
It is not the ’same’ thing yet, since you have km and m on the left side. That’s why you need to express either the km in meters or the meters in kilometers. We’ll go for the first option:
300 m in 1 sec
250. 000 m in x sec
Now you can do cross-multiplication:
(top left side)*(bottom right side ) = (bottom left side)*(upper right side)
300 * x= 250. 000 * 1
x = 250. 000 / 300 = 833. 3 seconds = 833. 3 / 60 = 13. 9 minutes = 833. 3/3600 = 0. 23 hours
You can also just fill in the formula:
time = distance / speed = 250 km / 300 m/s = 250. 000 m / 300 m/s = 833. 3 seconds = 13. 9 minutes = 0. 23 hours
SKYDIVER
If you want to express the speed in meters/second, first try to calculate how many kilometers one falls in 1 second (and afterwards, you convert the kilometers to meters)
3 km in 15 seconds
x km in 1 second
3 * 1 = x * 15
x = 3 / 15 = 0. 2 km
Thus, 0. 2 km in 1 second = 200 meters in 1 second = 200m/s
Converting meters per second to km per hour. Please note that 1 hour = 3600 seconds. Therefore, 1 second is (1/3600)hours. Note also that 1 km = 1000 m, thus 1 meter = (1/1000) km, 200 meters are therefore (200/1000) = 0. 2 km
0. 2 km in (1/3600) hours
x km in 1 hour (i. e. x km/h)
0. 2 * 1 = x * (1/3600)
x = 0. 2 * 3600 = 720 km
Thus, 200 m/s is the same as 720 km/h