Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.
Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
<h2>
Its velocity when it crosses the finish line is 117.65 m/s</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = ?
Time, t = 6.8 s
Displacement, s = 1/4 mi = 400 meters
Substituting
s = ut + 0.5 at²
400 = 0 x 6.8 + 0.5 x a x 6.8²
a = 17.30 m/s²
Now we have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = ?
Time, t = 6.8 s
Acceleration, a = 17.30 m/s²
Substituting
v = u + at
v = 0 + 17.30 x 6.8
v = 117.65 m/s
Its velocity when it crosses the finish line is 117.65 m/s
this is basically the same as volume, no?
So, 5.345*4.128*3.859=85.145
Its the answer of B speed and velocity
