Answer:
The dog catches up with the man 6.1714m later.
Explanation:
The first thing to take into account is the speed formula. It is
, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is 
Now, the distance equation for the man would be:

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.






The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.
That value is used in the man's distance equation.


Finally, the dog catches up with the man 6.1714m later.
The answer is 8 because multiplying 7 and 8 is 56
Explanation:
Given that,
Distance, s = 47 m
Time taken, t = 8.6 s
Final speed of the truck, v = 2.3 m/s
Let u is the initial speed of the truck and a is its acceleration such that :
.............(1)
Now, the second equation of motion is :

Put the value of a in above equation as :




u = 8.63 m/s
So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.
Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

And the distance traveled downwards is:

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

Replacing into the first equation:

Rationalizing:

Let's call v1 the final speed of the package dropped from a height H. Thus:

Let v2 be the final speed of the package dropped from a height 4H. Thus:

Taking out the square root of 4:

Dividing v2/v1 we can compare the final speeds:

Simplifying:

The final speed of the second package is twice as much as the final speed of the first package.