A man works on a striaght road from his home to market 2.5km away dwith a speed of 5km/h.Finding the market closed, he instantly
1 answer:
Answer:
5.625km/h
Explanation:
We are given that
Distance between home and market, d=2.5 km
Speed, v1=5km/h
Speed, v2=7.5 km/h
We have to find the average speed of the man over the interval of time 0 to 40min.
Time,
Using the formula
1 hour =60 min
Distance traveled by man in 10 min with speed 7.5 km/h=
Therefore,
Total distance covered=2.5+2.5/2=3.75 km
Time=40 min=40/60=2/3 hour
Average speed=
Average speed=
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We can solve for the acceleration by using a kinematic equation. First we should identify what we know so we can choose the correct equation.
We are given an original velocity of 24 m/s, a final velocity of 0 m/s, and a time of 6 s. We and looking for acceleration (a) in m/s^2.
The following equation has everything we need:
So plug in the known values and solve for a:
0 = 24 + 6a
-24 = 6a
a = -4 m/s^2
We are given an original velocity of 24 m/s, a final velocity of 0 m/s, and a time of 6 s. We and looking for acceleration (a) in m/s^2.
The following equation has everything we need:
So plug in the known values and solve for a:
0 = 24 + 6a
-24 = 6a
a = -4 m/s^2
<u>Explanation:</u>
The parametric representation of a line segment joining the points (a,b,c) and (l,m,n) is
r(t) = (1-t) . (a,b,c) + t . (l, m, n) where t ∈ |0, 1|
So, the parametric representation of a line segment joining the points (3,0,0) and (3,2,5) is
r(t) = (1 - t) . (3,0,0) + t . (3,2,5) where t ∈ |0, 1|
r(t) = (3(1 - t), 0, 0) + (3t, 2t, 5t) where t ∈ |0, 1|
r(t) = (3, 2t, 5t)
Given:
dr = (0, 2, 5) dt
Substitute = 9 + 29t² = u, 92tdt = du
Limit changes from 0→1 to 9 → 38
On solving this, we get:
Therefore, work done is