A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud
e of the total displacement, in km?
B. What is the direction of the displacement, measured in degrees west of north?
B. What is the direction of the displacement, measured in degrees west of north?
1 answer:
Answer:
Explanation:
given,
cyclist ride 6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.
vertical component = 9.28 cos∅
= 9.28 cos 27.27°
= 8.24 km
horizontal axis component = 9.28 sin ∅
= 9.28 sin 27.27°
= 4.5 km
distance of the final point from the origin
= 7.99 -(6.2-4.5)
= 6.29 km
displacement
d = 10.37 km
b)
θ = 37.36°
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Answer:
Explanation:
Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K
mg=k x
1 x 32= k x 2
K=16
And also give that damping force is 8 times the velocity so damping constant C=8.
We know that equation for spring mass system
my''+Cy'+Ky=F
Now by putting the values
1 y"+8 y'+ 16y=6 cos 4 t ----(1)
The general solution of equation Y=CF+IP
Lets assume that at steady state the equation of y will be
y(IP)=A cos 4t+ B sin 4t
To find the constant A and B we have to compare this equation with equation 1.
Now find y' and y" (by differentiate with respect to t)
y'= -4A sin 4t+4B cos 4t
y"=-16A cos 4t-16B sin 4t
Now put the values of y" , y' and y in equation 1
1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t
So by comparing the coefficient both sides
-16A+32B+16A=0 So B=0
-16 B-32 A+16B=6 So A=-3/16
y=-3/16 cos 4t
Now to find the CF of differential equation 1
y"+8 y'+ 16y=6 cos 4 t
Homogeneous version of above equation
So
So the general equation
Given that t=0 Y=0 So
t=0 Y'=0 So
The above equation is the general equation for motion.