Respuesta :
The question is incomplete. Here is the complete question.
The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined for this image, so the cars you see are in fact the same car, but photographed at differene times.
Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?
Answer: v = 6.5 m/s
Explanation: The question asks for velocity of the car. Velocity is given by:
[tex]v=\frac{\Delta x}{\Delta t}[/tex]
The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:
Δx = 7(5.3)
Δx = 37.1 m
The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:
1.3 frames = 1 s
7 frames = Δt
Δt = 5.4 s
Then, the car was driving:
[tex]v=\frac{37.1}{5.4}[/tex]
v = 6.87 m/s
The car drove at, approximately, a velocity of 6.87 m/s
The velocity of the car will be 6.5 m/s.The rate of change of displacement is defined as speed.
What is velocity?
The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
t is the time for camera deliver= 1.3 frames per second
l is the length = 5.3 meters
The instantaneous velocity is given as;
[tex]\rm v = \frac{\triangle x }{\triangle t } \\\\ \rm \triangle x = 7 \times 5.3 \\\\ \rm \triangle x = 37.1 m[/tex]
The time engaged is find as;
1.3 frames = 1 s
[tex]\rm \triangle t= 7 \ frames \\\\ \rm \triangle t=5.4 sec[/tex]
Hence the velocity of the car driving;
[tex]\rm v= \frac{37.1}{5.4} \\\\ \rm v= 6.87 m/sec[/tex]
Hence the velocity of the car will be 6.5 m/s.
To learn more abouty the velocity refe to the link;
https://brainly.com/question/862972
