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Expert: - 11/20/2009


Question
1. Intergrate with respect to x :(x^2-3) 2. The linear speed of a belt that passes round a pulley of radius 150mm is 20m/sec. If there  is no slip , how many revolutions per second are made by the wheel?      3.An object is thrown vertically upwards with a velocity of 8m/sec. What will be the greatest height reached and the time taken to reach that height. Take gravity as 10m/sec/sec 4Find the area enclosed by the curve y=3 cos 1/2x from x=0 to x= 3py/2.

Answer
1. ∫(x^2-3)dx

The answer would be to add one to the exponent and divide by new exponent.
That is, x^3/3 – 3x + C.


2. The linear speed of a belt that passes round a pulley of radius 150mm is 20m/sec.

If there is no slip , how many revolutions per second are made by the wheel?
The radius is 150 mm, or 015 m.

This makes the circumference 2πr, with r = 0.15, which is 0.3π = 0.9425,
almost a meter, in one revolution.

At 20 m/s, take 20/0.9425 to get revolutions.


3. An object is thrown vertically upwards with a velocity of 8m/sec.
What will be the greatest height reached and the time taken to reach that height.
Take gravity as 10m/s˛.

The equation would be to find where vt-at˛/2 = 0 with v as 8 and a as 10.

That is, f(t) = 8t-10t˛.  This means that f’(t) = 8 – 20t, and that needs to be set to 0.

That gives 8 – 20t = 0, or 20t = 8, or 5t = 2, or t = 2/5 = 0.4.

Putting that back into the height equation gives us
f(0.4) = 8(0.4) – 10(0.16) = 3.2 – 1.6 = 1.6 m.


4. Find the area enclosed by the curve y = 3•cos(x/2) from x=0 to x= 3π/2.

We have to integrate, and so ∫3cos(x/2)dx =6•sin(x/2).
Evaluate from 3π/2 down to 0 and get 6•sin(3π/4) - 6•sin(0) = 6(√2/2) = 3√2.

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