Question
2. For an object whose velocity in ft/sec is given by v(t) = -t^2 + 6, what is its displacement, in feet, on the interval t = 0 to t = 3 secs?
3. Find the velocity, v(t), for an object moving along the x-axis if the acceleration, a(t), is a(t) = cos(t) - sin(t) and v(0) = 3.
v(t) = sin(t) + cos(t) + 3
v(t) = sin(t) + cos(t) + 2
v(t) = sin(t) - cos(t) + 3
v(t) = sin(t) - cos(t) + 4
4. A pitcher throws a baseball straight into the air with a velocity of 72 feet/sec. If acceleration due to gravity is -32 ft/sec2, how many seconds after it leaves the pitcher's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.
2.25
2.5
4.25
4.5
I'm 90% sure that #2 is -3.00, #3 and #4 im lost. I think #4 is 2.25 but don't know #3
3. Find the velocity, v(t), for an object moving along the x-axis if the acceleration, a(t), is a(t) = cos(t) - sin(t) and v(0) = 3.
v(t) = sin(t) + cos(t) + 3
v(t) = sin(t) + cos(t) + 2
v(t) = sin(t) - cos(t) + 3
v(t) = sin(t) - cos(t) + 4
4. A pitcher throws a baseball straight into the air with a velocity of 72 feet/sec. If acceleration due to gravity is -32 ft/sec2, how many seconds after it leaves the pitcher's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.
2.25
2.5
4.25
4.5
I'm 90% sure that #2 is -3.00, #3 and #4 im lost. I think #4 is 2.25 but don't know #3
Answers
9.000
-9.000
10.596
-3.00
Those are the answer choices for #2
The concept you are dealing with has to do with distance (s), velocity (v), and acceleration (a)
You hopefully learned that
ds/dt = velocity
and dv/dt = acceleration
in #2
v(t) = -2t + 6
then s(t) = -t^2 + 6t + c
so s(3) = -9 + 18 + c = 9+c
and s(0) = 0+0+c
displacement = 9+c - c
= 9
#3
a(t) = cost - sint
then v(t) = sint + cost + k
v(0) = sin0 + cos0 + k = 3
0+1+k=3
k = 2
v(t) = sint + cost + 2 , which is choice #2
#4
you should be able to find
s(t) = -16t^2 + 72t as the distance equation.
the vertex of this downwards parabola gives you the answer.
s(t) = -16t(t - 4.5)
the x-intercepts are 0 and 4.5, with the vertex half-way between them or t = 2.25 seconds
Your were right on that one
You hopefully learned that
ds/dt = velocity
and dv/dt = acceleration
in #2
v(t) = -2t + 6
then s(t) = -t^2 + 6t + c
so s(3) = -9 + 18 + c = 9+c
and s(0) = 0+0+c
displacement = 9+c - c
= 9
#3
a(t) = cost - sint
then v(t) = sint + cost + k
v(0) = sin0 + cos0 + k = 3
0+1+k=3
k = 2
v(t) = sint + cost + 2 , which is choice #2
#4
you should be able to find
s(t) = -16t^2 + 72t as the distance equation.
the vertex of this downwards parabola gives you the answer.
s(t) = -16t(t - 4.5)
the x-intercepts are 0 and 4.5, with the vertex half-way between them or t = 2.25 seconds
Your were right on that one
Related Questions
The graph shows an object’s distance traveled over a period of time. Which statement is true? (1 poi...
When velocity is positive and acceleration is negative, what happens to the object's motion? (AKS 1...
Choose the correct two pieces of data to calculate the momentum of an object.
the mass of the objec...
Question What is the momentum of an object?(1 point) Responses the velocity of the object the veloc...