University EAS 208 (Fall 2019) Homework Assignment #1 Solution

This document provides a comprehensive solution to Homework Assignment #1 for the EAS 208 course, focusing on mechanical engineering principles. The assignment includes solutions to six problems covering topics such as particle acceleration, velocity, and displacement, analyzing motion from given velocity-time graphs, and projectile motion calculations. The solutions involve applying kinematic equations and calculus to determine velocity, position, and acceleration at different points, as well as calculating initial velocity and maximum height in projectile motion scenarios. The document also includes the analysis of motion with constant acceleration and the application of tangential and normal components of acceleration. Overall, the document offers a detailed, step-by-step approach to solving the problems, making it a valuable resource for students studying mechanical engineering.

HOMEWORK ASSIGNMENT #1
HOMEWORK ASSIGNMENT #1
Name of the Student
Name of the University
Author Note

HOMEWORK ASSIGNMENT #1
Q1-1:
Acceleration of the particle is given by,
a=−(0.1+sin ( s
0.8 )) m
s2
Here, s is the displacement.
Now, velocity of the particle is given by,
v(s)= −(0.1+sin ( s
0.8 ) ) ds
v(s) = -0.1s + 0.8cos(s/0.8) + c
Given, v(1) = 1 when s = 1.
=> 1 = -0.1*1 + 0.8cos(1/0.8) + c
=> c = 1+ 0.1 – 0.8cos(1/0.8) = 0.84
Hence, v(s) = -0.1s + 0.8cos(s/0.8) + 0.84.
b) The position where the velocity is maximum there its derivative, acceleration is 0.
−(0.1+sin ( s
0.8 )) = 0
=> sin ( s
0.8 ) = -0.1
=> (s/0.8) = arcsin(-0.1)
=> s = -0.1001*0.8 = -0.08
Now, this is to be confirmed when a’(s) < 0 or v’’(s) at s = -0.08 < 0

HOMEWORK ASSIGNMENT #1
a' ( s )=− ( 1
0.8 )cos ( s
0.8 )¿
a' (−0.08 ) = −( 1
0.8 )cos (−0.08
0.8 )¿ < 0
Hence, v(s) is maximum at s = -0.08.
c) The maximum velocity v(-0.08) = -0.1*(-0.08) + 0.8cos(-0.08/0.8) + 0.84 = 1.644 m/sec.
Q1-2:
v(t) curve is given below.
v(t) = 3t 0 <=t <= 4
= 12 4<t<=8
= (¾)t + 6 8<t<=12
For s-t curve:
s(t) = 3t^2/2 + c = 3t^2/2 (as at t = 0, s = 0) 0 <=t <= 4
= 12t + c => 24 = 12*4 + c => c = -24
= 12t -24 (as at t = 4, s = 24) 4<t<=8

HOMEWORK ASSIGNMENT #1
= (¾)t^2/2 + 6t + c => 72 = (3/4)*64/2 + 6*8 + c => c = 72 – 48 – 24 = 0
= (3/8)t^2 + 6t
For a-t curve:
a(t) = 3 0 <=t <= 4
= 0 4<t<=8
= ¾ 8<t<=12
0 2 4 6 8 10 12 14
0
20
40
60
80
100
120
140
a-t and s-t curve in time [0,12]
a(t) s(t)
Time in sec
a in m/s^2 and s in m
Q1-3:
Given, the angle to the horizon θ = 25°.
At the distance where the ball touches fence x range is 110 m and y distance is 3 m.
i.
Now, y displacement at time t is

HOMEWORK ASSIGNMENT #1
y(t) = y0 + v0*t*cos(θ) – (½)g*t^2 => 3 = 1 + v0*t*cos(25) – (½)*9.8*t^2
x(t) = v0*t*cos(θ) => 110 = v0*t*cos(25) => t = 110/(v0*0.906) = 121.41/v0
v0*t*cos(25) – (½)*9.8*t^2 = 2
 121.41cos(25) – 4.9*14740.388/v0^2 = 2
 4.9*14740.388/v0^2 = 121.41cos(25) – 2 = 108.03
 v0 = 25.857 m/sec.
ii.
The maximum height h is
h= y 0+v 02∗sin ( θ )
2 g = 1 + 25.857^2*sin(25)/(2*9.8) = 15.416 m.
iii. Now, when the ball reaches the fence the x displacement is 110 m.
110 = 25.857*t*cos(25) => t = 110/(25.857*cos(25)) = 4.694 secs.
Q1-4:
y(t) = y0 + v0*t*cos(α) – (½)g*t^2
x(t) = v0*t*cos(α)
Given, the inclination angle is 25 degrees, v0 = 5 √2 ft/s
Now, tan^(-1) (y/x) = 25 degrees
 tan(25) = (v0*sin(α)t – 0.5gt^2)/(v0*cos(α)t)
 (1/2) gt + cos(α) tan(25°) - 5 √2sin(α) = 0
 4.9t + cos(α) tan(25°) - 5 √2sin(α) = 0 (1)
0.9 = 5 √2*t*cos(α) – (½)g*t^2 (2)

HOMEWORK ASSIGNMENT #1
Solving the above two equations gives α = 2.08 rad = 119 degrees.
Hence, the particle hit the surface at a height of 0.9 ft with an angle of 119 degrees.
Q1-5:
Given initial speed of the car u = 50 mi/h.
Final speed of car = 65 mi/h and the speed is increased with a constant acceleration.
After 2 seconds v = 50 + 2f => 65 = 50 + 2f => f = 15/2 = 7.5 mi/hr^2
Hence, the acceleration after 2 seconds of passing the point A will be 7.5 mi/hr^2.
Q1-6:
a)
The ball is rolling on the surface with equation f(x) = 2x^2 – 4x + 5.
f(4) = 2*4^2 – 4*4 + 5 = 2*16 – 16 + 5 = 32 – 16 + 5 = 21.
Hence, angle at x=4 => tan^(-1)(21/4) = 1.382 rad.
Now, the angle between normal and the tangential component = pi – (pi/2 + 1.382) = 0.189
rad = 10.828 degrees.
Hence, the tangential component of acceleration at point A = 4 m/s2.
Now, the normal component is 4*cos(10.828) = 3.92 m/sec^2
b)
The angle between the acceleration and velocity vectors at A is given by,
θ at x =4=tan−1 velocity
accleareition = tan−1 5
4 = 0.896 rad = 51.33 degrees.

HOMEWORK ASSIGNMENT #1