An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is s = 6 cos(t) + 5 sin(t), t ≥ 0, where s is measured in centimeters and t in seconds. (Take the positive direction to be downward.) (a) Find the velocity and acceleration at time t.

V = 1.4m/a

acceleration = - 7.68m/s^2

Step-by-step explanation: The question can be solved by using differential calculus.

Please find the attached file for the solution

Velocity = -7.806 m/s and acceleration = -0.253 m/s²

Step-by-step explanation:

From the question, we know that;

s = 6 cos(t) + 5 sin(t)

Velocity (V(t)) =S'(t) =-6sin(t) + 5cos(t)

Acceleration (a(t)) = V'(t) = - 6cos(t) - 5 sin(t)

A) Hence, we have to solve for s =0;

0 = 6 cos(t) + 5 sin(t)

Thus, 5 sin(t) = - 6 cos(t)

Sin(t) = -(6/5)cos(t)

Sin(t) = - 1.2Cos(t)

Sin(t)/Cos(t) = - 1.2

Tan(t) = - 1.2

t = Tan^(-1)(-1.2)

t = -51.9

But this problem is in radians and thus, converting to radians we get;

t = -51.9 x 0.0175 = - 0.9083 radians

we know t > 0. The first positive solution to tan (t) will be = - 0.9083 + π = - 0.9083 + 3.1416 = 2.2333 seconds

A) Velocity = -6sin(t) + 5cos(t)

Thus, since t = 2.2333 and using radians calculator;

Velocity = -6sin(2.2333) + 5cos(2.2333) = -4.731 - 3.075 = -7.806 m/s

B) Acceleration = -6cos(t) - 5 sin(t)

At t = 2.2333, and using radians calculator

Acceleration = -6cos(2.2333) - 5 sin(2.2333) = 3.689 - 3.942 = -0.253 m/s²

v = 1/3 a^2h

step-by-step explanation:

if the height is 57.7cm and the height of the slant edge is 18.2cm, then the volume is 1/3 a^2h.

according to egenuity the answer is

step-by-step explanation:

d. 344.22