Solved Assignments and Study Material for Desklib

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