## Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

# Chapter 12 - Kinematics of a Particle - Section 12.6 - Motion of a Projectile - Problems - Page 51: 85

#### Answer

$d=204$ m $v=41.8$ m/s $a=4.66$ m/s$^2$

#### Work Step by Step

First, we solve for the distance as follows: $x=2t^2$ $y=0.04t^3$ At $t=10$s, $x=200$m, $y=40$m $d=\sqrt{200^2+40^2}=204$ m $v_x=\frac{dx}{dt}=4t$ $a_x=\frac{dv_x}{dt}=4$ $v_y=\frac{dy}{dt}=0.12t^2$ $a_y=\frac{dv_y}{dt}=0.24t$ Now, find the velocity and acceleration at $t=10$ s, $v=\sqrt{40^2+12^2}=41.8$ m/s $a=\sqrt{4^2+2.4^2}=4.66$ m/s$^2$

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