{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW2 - schneiter(tcs639 HW02 Ross(89152 This print-out...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
schneiter (tcs639) – HW02 – Ross – (89152) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points The position of a softball tossed vertically upward is described by the equation y = c 1 t - c 2 t 2 , where y is in meters, t in seconds, c 1 = 8 . 62 m / s, and c 2 = 2 . 77 m / s 2 . Find the ball’s initial speed v 0 at t 0 = 0 s. Correct answer: 8 . 62 m / s. Explanation: Basic Concepts: v = dx dt a = dv dt = d 2 x dt 2 Solution: The velocity is simply the derivative of y with respect to t : v = dy dt = 8 . 62 m / s - 2(2 . 77 m / s 2 ) t, which at t = 0 is v 0 = 8 . 62 m / s . 002 (part 2 of 3) 10.0 points Find its velocity at t = 0 . 691 s. Correct answer: 4 . 79186 m / s. Explanation: Substituting t = 0 . 691 s into the above formula for v , we obtain v = 8 . 62 m / s - 2(2 . 77 m / s 2 )(0 . 691 s) = 4 . 79186 m / s . 003 (part 3 of 3) 10.0 points Find its acceleration at t = 0 . 691 s. Correct answer: - 5 . 54 m / s 2 . Explanation: The acceleration is the derivative of velocity with respect to time: a = dv dt = - 2(2 . 77 m / s 2 ) = - 5 . 54 m / s 2 .
Background image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}