a) From the diagram using either scale drawing or some trig. The displacement is 500km 53 degrees south of east.
b) average speed = total distance gone / time taken
average speed = (300 + 400)km / 2hours = 700/2 = 350km/h
c) average velocity = displacement / time = 500km / 2hours = 250km/h
Average value for g = (9.72 + 9.83 + 10.52 + 9.67 + 9.82) / 5 = 9.91m/s2
The random error = (max reading - min reading)/ number of readings
Random error = (10.52 - 9.67) / 5 = 0.17
So overall Linda would state her result for g as 9.91m/s2 plus or minus 0.17m/s2
In all these problems always mark down all the data like this:
u, initial velocity = 12m/s
v, final velocity = ?
a, acceleration = 3m/s2
S, displacement = 200m
t, time = ?
From the data we can use u, a, S and we need to find v.
So it looks that we will have to use v2 = u2 + 2aS
So plugging in the values gives v2 = 122 + (2x3x200)
v2 = 1344
v = 36.7m/s
The graph shows a constant acceleration so each second the velocity will increase by 3m/s...check the graph below:
a) Once again in this type of problem think it through in pictures. The player leaves the ground with a vertical velocity and this velocity is reduced to zero under gravity over a height of 0.85m. We want to find his initial vertical velocity.
So start with a list of what you know and have to find.
0m/s, because at the top vertical velocity = 0m/s
-9.8m/s remember acceleration of gravity acts downwards
0.85m the height the player jumps .
To find u we can use the equation v2 = u2 + 2aS
So plugging in the numbers gives:
02 = u2 + (2 x -9.8 x 0.85)
Rearranging gives u2 = 16.7
So u = 4.1m/s
b) The time he was in the air can be found the same way. We will calculate the time it takes for the player to reach his maximum height and assuming no air resistance the total time he spent in the air will be double this answer. We have:
u = 4.1m/s
v = 0m/s
a = -9.81m/s2
S = 0.85m
t = ?
Looking at the list of data we have: u,v,a and we need t. So it looks like we will use the formula:
v = u + at
0 = 4.1 + (-9.81 x t )
Rearranging gives t = 4.1 / 9.81 = 0.42s.
Therefore the player was in the air for 2 x 0.42s = 0.84s.
c) The horizontal distance = horizontal velocity x time in air.
Horizontal distance = 2.5m/s x 0.84s