## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.4 - The Derivative as a Rate of Change - Exercises - Page 145: 19

#### Answer

a) It takes the balls $t=4/7$ seconds to fall the first $160cm$. The average velocity for this period is $280cm/sec$. b) The speed of the balls is $560cm/sec$. Their acceleration is $980cm/sec^2$. c) Light flashing reaches as fast as $30$ flashes per second.

#### Work Step by Step

$$s=490t^2$$ a) The time $t$ it takes to fall the first $160$ cm, which is $s=160$, is $$t^2=\frac{s}{490}=\frac{160}{490}=\frac{16}{49}$$ $$t=\frac{4}{7}(sec)$$ (the time $t$ is never negative) The average velocity for this period is $$\frac{s}{t}=\frac{160}{\frac{4}{7}}=280(cm/sec)$$ b) Find instantaneous velocity $v$ and instantaneous acceleration $a$: $$v=\frac{ds}{dt}=(490t^2)'=980t$$ $$a=\frac{dv}{dt}=(980t)'=980$$ The balls reached their $160$-cm mark at $t=4/7sec$. The speed of the balls at that moment is $$|v(4/7)|=|980\times(4/7)|=560(cm/sec)$$ Since $a=980cm/sec^2$, it is a constant. So the acceleration of the balls at $t=4/7sec$ is $980cm/sec^2$. c) In the photograph, from position $65cm$ to position $90cm$, not counting the first image of the balls' position at $s=65cm$, the positions of the balls were captured twice by the machine. In other words, 2 light flashes were produced during the time the balls fell from $s=65cm$ to $s=90cm$. Now, to know how fast the light flashing is, we need to calculate how long it takes the balls to fall from position $s=65cm$ to position $s=90cm$. - For $s=65cm$: $t=\sqrt{\frac{65}{490}}\approx0.364sec$ - For $s=90cm$: $t=\sqrt{\frac{90}{490}}\approx0.429sec$ So it takes approximately $0.429-0.364=0.065sec$ to fall between these two position. Since there are 2 flashes in this $0.065sec$, in a second, the number of flashes are $$\frac{2}{0.065}\approx30(flashes)$$ So light flashing can reach $30$ flashes per second.

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