# HW3.sol - homework 03 – RAHMAN TARIQUE – Due Feb 4 2008...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: homework 03 – RAHMAN, TARIQUE – Due: Feb 4 2008, 3:00 am 1 Question 1, chap 2, sect 6. part 1 of 1 10 points A toy rocket, launched from the ground, rises vertically with an acceleration of 20 m / s 2 for 9 . 3 s until its motor stops. The acceleration of gravity is 9 . 8 m / s 2 . Disregarding any air resistance, what max- imum height above the ground will the rocket achieve? Correct answer: 2 . 63 km (tolerance ± 1 %). Explanation: Basic Concepts: v = v + at y = v t + 1 2 at 2 since the initial position is zero. Solution: While the motor is running, the rocket rises to a height of h 1 = 1 2 a t 2 = 1 2 ( 20 m / s 2 ) ( t ) 2 · 1 km 1000 m = 0 . 8649 km When the motor stops running, the speed of the rocket is v 1 = at At this point in time the rocket continues to rise, but now its acceleration (deceleration) is due to gravity alone, and its initial velocity is v 1 = at . At its maximum height its velocity is zero, so v 2 = 0 = v 2 1 − 2 g h 2 and 2 gh 2 = v 2 1 h 2 = v 2 1 2 g = ( at ) 2 2 g = ( 20 m / s 2 ) 2 (9 . 3 s) 2 2 (9 . 8 m / s 2 ) = 1 . 7651 km Thus the total height obtained by the rocket is h = h 1 + h 2 = 0 . 8649 km + 1 . 7651 km = 2 . 63 km Question 2, chap 2, sect 7. part 1 of 1 10 points An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the acceleration a of the object is given by a = g − b v , where v is the object’s speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time? 1. v = ( g + a ) t b 2. v = ( g e bt ) b 3. v = v + g t, v negationslash = 0 4. v = g t − b t 2 5. v = g ( 1 − e − bt ) b correct Explanation: At time t = 0, the speed of the object is zero, and at time t = ∞ , the acceleration is zero, corresponding to a speed v = g b . Check the five choices, and it shows that the only possible answer is v = g ( 1 − e − bt ) b . Note: The answer can be directly obtained by integration: a = g − b v d v dt = − g parenleftbigg b v g − 1 parenrightbigg homework 03 – RAHMAN, TARIQUE – Due: Feb 4 2008, 3:00 am 2 integraldisplay v dv b g v − 1 = − g integraldisplay t dt g b ln parenleftbigg b g v − 1 parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle v = − g t vextendsingle vextendsingle vextendsingle vextendsingle t g b bracketleftBig ln parenleftbigg b g v − 1 parenrightbigg − ln( − 1) bracketrightBig = − g t ln parenleftbigg 1 − b g v parenrightbigg = − b t 1 − b g v = e − b t b g v = 1 − e − b t v = g b parenleftBig 1 − e − b t parenrightBig . Question 3, chap 2, sect 7....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

HW3.sol - homework 03 – RAHMAN TARIQUE – Due Feb 4 2008...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online