Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 13 - Section 13.4 - Motion in Space: Velocity and Acceleration - 13.4 Exercise - Page 879: 31

Answer

$(250, -50,0)$ ; $10\sqrt{93} m/s \approx 96.4 ft/s$

Work Step by Step

Use velocity-time equation. $v=u+at$ $v=u+(\lt0,-4,-32 \gt \lt 50,0,80 \gt)t$ $v=\lt 50,-4t,80-32t \gt$ $r(t)=\int v(t)=\lt 50t,-2t^2,80t-16t^2 \gt$ Since, $80t=16t^2=0$ $t=5$ Thus, $r(5)=\lt 50(5),-2(5)^2,80(5)-16(5)^2 \gt=\lt 250, -50,0 \gt$ $v(5)=\lt 50, -20,-80 \gt$ Thus, final speed is $|v(5)|=\sqrt {(50)^2+( -20)^2+(-80)^2 }=\sqrt {100+400+6400}=10\sqrt{93} m/s \approx 96.4 ft/s$
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