Answer:
The runner's acceleration was
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
Solving for a:
The runner speeds up from vo=5 m/s to vf=9 m/s in t=4 seconds, thus:
The runner's acceleration was
A) -2.0 m
Look at the ray diagram attached in the picture, where:
p identifies the location of the object
q identifies the location of the image
F identifies the focus of the mirror
Each tick represents 1 m
We have
p = 6.0 m is the distance of the object from the mirror
f = -3.0 m is the focal length
From the ray diagram, we see that q has a distance of 2.0 m from the mirror, and it's on the other side of the mirror compared to the object, so
q = -2.0 m
This can also be verified by using the mirror equation:
B) Upright and virtual
As we see from the picture, the image is upright, since it has same orientation as the object.
Also, we notice that the image is on the other side of the mirror, compared to the object. For a mirror,
- An image is said to be real if it is on the same side of the object
- An image is said to be virtual if it is on the opposite side of the mirror
Therefore, this means that the image is virtual.
