HELP: Suppose that an object moves along the y -axis so that its location is y=x^2+4x at time x .?

Question

Suppose that an object moves along the y -axis so that its location is y=x2+4x at time x . (Here y is in meters and x is in seconds.)
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I've tried this problem 28 times!! PLEASE HELP!

Find the average velocity for x changing from 5 to 5+h seconds.

Answer 1

1) The velocity, v(x) = dy/dx = 2(x+2);

2) By definition, the average velocty, q, in the closed interval [5, 5+h], is (1/h) times the definite integral of v(x) from 5 to 5 +h

3) After some effort; q = (1/h){ (5+h)^2 + 4(5+h) - 45 } = h + 14;

Answer 2

The velocity is the derivative

V = dy/dx= 2x+4 m/s

At time 5 its velocity is 14m/s

At time mx =5+h its velocity is 2(5+h) +4= 14 +2h

distance ttravelled in time h from t=5 would be

(5+h)^2 +4*(5+h) - 5^2-4*5= 25 +10h +h^2 +20+4h -25-20=

h^2 +14h
The mean velocity is distance/time = (h^2+14h)/h=h+14 m/s

Answer 3

Do the differntiation of given action and find out the velocity dy/dx

y = x^2+4x
dy/dx = 2x+4
X = (x1,x2) = (5,5+h)

Put this limit values in dy/dx

velocity will be = [{2*(5+h)+4} - (2*5+4)] = 2h
Place the value of h and get the answer