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# calculus question

Assume that∫x^a dx=x^(a+1)/(a+1)+C for all real numbers a≠-1.

Take g=9.8ms-2

1. Find the trajectory of a particle, s(t), from the given information:

a(t)=t+1, v(1)=2, s(3)=4

2. Evaluate the following integrals by the method of substitution:

(a) ∫cos 2πt dt

(b) ∫x^3√(x^4 -1) dx

(c) ∫x√(x+1) dx

(d) ∫(sec^2)2xtan2x dx (d/dx secx=secxtanx)

3. An object fall from a height of 30km above the ground. Suppose air

resistance and all other forces except gravity can be ignored. Find the speed of the object when it reaches the ground.

4. A rocket is launched vertically upwards with a constant acceleration of

5ms-2. Tthe engine is shut down after 20s. Find the maximum height(from

ground) reached by the rocket.(Neglect air resistance)

5. An object at ground level is projected with an initial speed of 30ms-1 at an

elevation angle of 60 degrees to the horizontal. Find its range on a plane

inclined at an angle of 30 degrees to the horizontal.

### 1 Answer

- Audrey HepburnLv 71 decade agoFavorite Answer
1. a(t) = t + 1

v(t) = ∫a(t)dt = t2/2 + t + C

v(1) = 2

So, 2 = 1/2 + 1 + C, C = 1/2

v(t) = t2/2 + t + 1/2

s(t) = ∫v(t)dt = t3/6 + t2/2 + t/2 + C'

s(3) = 4

So, 4 = (3)3/6 + (3)2/2 + 3/2 + C', C' = -13/2

The trajectory: s(t) = t3/6 + t2/2 + t/2 - 13/2

2.a. ∫cos2πt dt = 1/2π ∫cos2πt d(2πt) = sin2πt / 2π + C

b. Let u = x4 - 1, du = 4x3dx

∫x3√(x4 -1) dx = 1/4 ∫√u du = 2u3/2/3 + C = 2(x4 - 1)3/2 / 3 + C

c. Let u = x + 1, du = dx

∫x√(x+1) dx = ∫(u - 1)√u du = ∫(u3/2 - u1/2) du

= 2u5/2 / 5 - 2u3/2 / 3 + C = 2(x + 1)5/2 / 5 - 2(x + 1)3/2 / 3 + C

d. ∫sec22xtan2x dx = 1/2 ∫sec2x d(sec2x) = sec22x / 4 + C

3. In fact, this question cannot simply use the equation of motion v2 = u2 - 2as. We should use integration to tackle the problem. However, here I will use both methods, and you can compare the answers.

By equation of motion v2 = u2 + 2as

v2 = 0 + 2(10)(30 X 103)

Speed, v = 775 ms-1

Method of integration:

圖片參考：http://i295.photobucket.com/albums/mm158/Audrey_he...

4. By v = u + at

v = 0 + 5(20)

v = 100 ms-1

By s = ut + 1/2 at2

s = 0 + 1/2 (5)(20)2

s = 1000 m

After shutting the engine, by v2 = u2 + 2as

0 = (100)2 - 2(10)s

s = 500 m

So, the maximum height from ground of the rocket is 1000 + 500 = 1500 m

5.

圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_19...

Source(s): Myself~~~- Login to reply the answers