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
a) 10 m/s
b) 25 m
Explanation:
a)
The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:
where
u is the initial velocity
v is the final velocity
a is the acceleration
t is the time
For the body in this problem:
u = 0 (the body starts from rest)
is the acceleration
t = 5 s is the time
So, the final velocity is
b)
In this second part, we want to calculate the distance travelled by the body.
We can do it by using another suvat equation:
where
u is the initial velocity
v is the final velocity
a is the acceleration
s is the distance travelled
Here we have
u = 0 (the body starts from rest)
is the acceleration
v = 10 m/s is the final velocity
Solving for s,