Question

Question 9 (1 point)

An airplane starts from rest (vo = 0) and accelerates down a runway at 2 m/s2 for 23 s until it takes off. Determine the distance traveled before takeoff.

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Question 10 (1 point)

An airplane starts from rest (vo = 0) and accelerates down a runway at 4 m/s2 and covers 245 m before it takes off. Determine the time it takes.

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Question 11 (1 point)

An airplane starts from 2 m/s and accelerates down a runway at 1 m/s2 and covers 446 m before it takes off. Determine the time it takes.

Note, we still need x = vo t + 1/2 a t2, but the formula will still have t and t2.

This is called a quadratic equation. Probably the best way to solve this is with the quadratic formula (your other choices are factoring, completing the square, graphing or numerical methods)

Steps:

Simplify as much as possible and arrange in descending order - that is, get everything on one side of the equation with a zero on the other side and rearrange the terms so that t2 comes first, t comes second and the constant term is lastthe quad formula has a, b, and c which are the coefficients of the t2 term, the t term and the constant term including the positive or negative signsexample:If 4 = 2 t + 1/2 (8) t2Simplify 1/2 (8) = 4, subtract 4 from both sides and rearrange the terms to get 0 = 4t2 + 2t - 4now a = 4, b = 2, c = -4plug these into the quadratic formula and get the two solutions, one when we use the + and the other when we use the - in the formula, just before the square root. Usually we want the positive solution(-2+√(22-4(4)(-4))).(2*4) = 0.78(-2-√(22-4(4)(-4))).(2*4) = -1.3

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Question 12 (1 point)

Sometimes you have to do the problem in two steps:

A car moving at 16 m/s comes to a stop in 6 s. Find the distance it covers.

Here we see that vO = 16 m/s, v = 0 m/s (comes to a stop), and t = 6 s, and they want x.

We don't have a formula that relates vO , v , t and x. So:

first use v = vO + at to find athen use x = vO t + 1/2 at2 to find x

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Question 13 (1 point)

If a car moving 13 m/s comes to a stop in 5 seconds, find the acceleration

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Question 14 (1 point)

A dolphin jumps straight up out of the water at a velocity of 9 m/s. How high

does his body rise above the water before he comes to a stop due to gravity pulling downward at 9.8 m/s2 ?

Reminder: vo is up and acceleration is down - in Physics opposite directions get represented by opposite signs, say, up is positive and down is negative.

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Answer 1

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