You have a distance and a time so you can work out the bikes speed which would be distance/time so 40mph.
Answer:
t = 3.516 s
Explanation:
The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:

Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0
If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.
This can be done knowing that
1 km = 1000 m
1 h = 3600 s
Therefore
1 km/h = (1000/3600) m/s = 0.2777... m/s
100 km/h = 27.777... m/s
Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:
27.777 m/s = 7.9 m/s^2 t
Solving for t
t = (27.7777 / 7.9) s = 3.516 s
Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>


, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so 
On Earth, gravity adds 9.8 m/s to the downwsrd speed of any falling object, every second. Beginning from zero downward speed, the speed grows to (8 x 9.8) = 78.4 m/s downward after 8 sec.
Explanation:
Could you also show the diagram