A) position time graph for both is shown
here one of the graph is of lesser slope which means it is moving with less speed while other have larger slope which shows larger speed
At one point they intersects which is the point where they both will meet
B) Let the two will meet after time "t"
now we can say that
if they both will meet after time "t"
then the total distance moved by you and other person will be same as the distance between you and home
so it is given as



so they will meet after t = 6 min
so from position time graph we can see that two will meet after t = 6 min where at this position two graphs will intersect
Answer:
Time, t = 12 minutes
Explanation:
It is given that,
A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. Let west direction is negative and east direction is positive. The displacement of the cyclist is :

d = 4800 m
Let us assumed that the average speed of the cyclist is, v = 24 km/h = 6.66667 m/s
Let t is the time taken by the cyclist to complete the trip. The velocity of an object is given by :



t = 719.99 seconds
t = 720 seconds
or
t = 12 minutes
So, the time taken by the cyclist to complete the trip is 12 minutes. Yes, the time taken by the cyclist to complete the trip is reasonable. Hence, this is the required solution.
Answer:
20.0 cm
Explanation:
Here is the complete question
The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Solution
Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.
Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.
Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m
Now, P' = 1/u + 1/v
1/u = P'- 1/v
1/u = 55.0 D - 1/0.02 m
1/u = 55.0 m⁻¹ - 1/0.02 m
1/u = 55.0 m⁻¹ - 50.0 m⁻¹
1/u = 5.0 m⁻¹
u = 1/5.0 m⁻¹
u = 0.2 m
u = 20 cm
So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.
Number 4 is c , number 5 is a , number 6 is d and 7 is a