(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
Explanation:
(a) Displacement of an object is the shortest path covered by it.
In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.
0.4 miles = 0.64 km
displacement = 0.7-0.3+0.64 = 1.04 km
(b) Average velocity = total displacement/total time
t = 15 min = 0.25 hour

Hence, this is the required solution.
Conservation of linear momentum:
m*v inital = m*v final
0.06*0.7 + 0.03*0 = 0.06*(-0.2) + 0.03*v
(my algebra, or use ur calculator: 0.06*.07=0.042, etc ... or ur teacher may think you got some help)
0.06*(0.7+0.2)=0.03*v, v = 0.06*0.9/0.03=1.8 m/s
Answer 1.8 m/s (positive, to the right).
Answer:
160 Hz , 240 Hz , 400 Hz
Explanation:
Given that
Frequency of forth harmonic is 320 Hz.
Lets take fundamental frequency = f₁

f₁=80 Hz
Frequency of first harmonic = f₂
f₂=2 f₁
f₂ =2 x 80 = 160 Hz
Frequency of second harmonic = f₃
f₃= 3 f₁=3 x 80 = 240 Hz
Frequency of fifth harmonic = f₅
f₅= 5 f₁= 5 x 80 = 400 Hz
Three frequencies are as follows
160 Hz , 240 Hz , 400 Hz