- An object is thrown in the air from an initial height of 12 feet, with an initial upward velocity of 16 feet/second?
How long will the object be in the air?
What will the velocity of the object be after 1 second? - An object is thrown in the air from an initial height of 48 feet, with an initial upward velocity of 32 feet/second?
How long will the object be in the air?
What will the velocity of the object be after 2 seconds? - An object is thrown in the air from an initial height of 100 feet, with an initial upward velocity of 10 feet/second?
How long will the object be in the air?
What will the velocity of the object be after 2 seconds? - An object is thrown in the air from an initial height of 400 feet, with an initial upward velocity of 50 feet/second?
How long will the object be in the air?
What will the velocity of the object be after 4 seconds? - An object is thrown in the air from an initial height of 75 feet, with an initial upward velocity of 200 feet/second?
How long will the object be in the air?
What will the velocity of the object be after 5 seconds? - Suppose it costs -x^2 + 400x dollars to produce x computers/day. Compute the marginal cost to estimate the cost of producing one more computer each day, if current production is 100 computers/day.
- If the position of an object at time t is given by the function
s(t) = 3t + 2 meters what are the velocity and acceleration when t = 3 seconds? - If the position of an object at time t is given by the function
s(t) = t3 - t meters what are the velocity and acceleration when t = 5 seconds? - If the position of an object at time t is given by the function
s(t) = 3t3 - 10t2 meters what are the velocity and acceleration when t = 2 seconds? - If the position of an object at time t is given by the function
s(t) = sin t meters what are the velocity and acceleration when t = pi/4 seconds? - If the position of an object at time t is given by the function
s(t) = cos 2t meters what are the velocity and acceleration when t = pi/4 seconds? - If the position of an object at time t is given by the function
s(t) = sin t + t meters what are the velocity and acceleration when t = pi/2 seconds? - If the position of an object at time t is given by the function
s(t) = cos t + sin t meters what are the velocity and acceleration when t = pi seconds? - If the position of an object at time t is given by the function
s(t) = (1/2)t3 + 2t meters what are the velocity and acceleration when t = 1 second? - If the position of an object at time t is given by the function
s(t) = 6t2 - 8t + 19 meters what are the velocity and acceleration when t = 4 seconds?
Saturday, March 21, 2009
Question set 6
Subscribe to:
Post Comments (Atom)
(15)
ReplyDeletes(t)= 6t2 - 8t- 19
velocity: ds/dt= 12t-8 m/s
ds/dt when t=4
ds/dt= 12(4) - 8
=48-8
=40
to find acceleration you would have to differentiate again
this will give you d^2s/dt2 = 12 m/s
13.
ReplyDeletes(t) = cos t + sin t
rewrite
s = (cos t)/t + (sin t)/t
now velocity is the same as rate of change which is ds/dt
ds/dt = (-sin t)/t + (cos t)/t
and acceleration is the same as rate of change of velocity which is differentiating ds/dt to give d^2(s)/dt^2
so d^2(s)/dt^2 = (-cos t)/t - (sin t)/t
therefore when t = 3.14
ds/dt = (-sin 3.14)/3.14 + (cos 3.14)/3.14
ds/dt = 0.3
and
d^2(s)/dt^2 = (-cos 3.14)/3.14 - (sin 3.14)/3.14
d^2(s)/dt^2 = -0.33
no 7
ReplyDeletes(t) = 3t + 2
velocity: ds/dt = 3 m/s
acceleration: d^2s/dt^2 = 0 m/s^2
no 8
ReplyDeletes(t) = t^3 - t
ds/dt = 3t^2 - 1
at t = 5 seconds
velocity = 3(5)^2 - 1 = 74 m/s
acceleration: d^2s/dt^2 = 6t
at t = 5 seconds
acceleration = 6*5 = 30 m/s^2
no 9
ReplyDeletes(t) = 3t^3 - 10t
ds/dt = 9t^2 - 20t
at t= 2 sec
velocity = 9(2)^2 - 20(2)
= -4 m/s
acceleration: d^2s/dx^2 = 18t-20
at t = 2 sec
acceleration = 18(2) - 20 = 16m/s^2
no. 1
ReplyDeletei'm not sure how to start this question but i will give it a try.
since velocity is the same as rate of change of distance w.r.t time, i will represent it by ds/dt
the object travelled from 12 to x so distance travelled will be x - 12
remembering the equation for velocity i can write a differentiated equation
ds/dt = (x - 12)/t
now when t= 1
ds/dt = x - 12
velocity is the same as rate of change of distance w.r.t time. jus a correction to the solution
ReplyDeleteQuestion #2:
ReplyDeleteAn object is thrown in the air from an initial height of 48 feet, with an initial upward velocity of 32 feet/second?
How long will the object be in the air?
Answer:
So, velocity implies rate of change!
Which implies; dh/dt.
But, when t = 0, h = 48
And, dh/dt = 32
therefore, dh = 32 dt
So, i'm guess there is the need to integrate!
14)If the position of an object at time t is given by the function
ReplyDeletes(t) = (1/2)t3 + 2t meters what are the velocity and acceleration when t = 1 second?
The velocity ........ ds/dt = 3/2t^2 + 2
Therefore ds/dt when t = 1
= 3/2 (1)^2 + 2
= 7/2 or 3 and a half.
The acceleration will therefor be
d^2s/dt2
= 2.25t
when sbstituting 1 for the value of t
= 2.25 * 1
= 2.25m/s
For question 15)
ReplyDeletes(t)= 6t2 - 8t- 19
therefore when differeniated
ds/dt= 12t-8 m/s
ds/dt when t=4
ds/dt= 12(4) - 8
=48-8
=40
Not sure how to find the acceleration but i think you will have to differeniate the equation again, Correct Me If Am Wrong Please.........
no. 11
ReplyDeletevelocity is the same as ds/dt and acceleration is the same as d^2(s)/dt^2.
also for y = sinx, dy/dx = cosx
for y = cosx, dy/dx = -sinx
therefore:
s(t)= cos 2t
substitute x = 2t
dx/dt = 2
rewrite: s(t) = cos x
ds/dx = -sin x
ds/dt = ds/dx * dx/dt
= -sin x * 2
= -2sin 2t
ds/dt = -2sin 2t
let x = 2t
dx/dt = 2
rewrite: ds/dt = -2sin x
ds/dtdx = -2cos x
hence
d^2(s)/dt^2 = ds/dtdx * dx/dt
= -2 cos x * 2
= -4 cos 2t
when t = pi/4
velocity = -2 sin [pi/4]
= -2 * 0.0137
= 0.0274
acceleration = -4 cos [pi/4]
= -4 * 0.9999
= -3.9996
hey,i am not sure about these velocity type questions either....but i did some research on the net.....!
ReplyDelete1.v^2=u^2+2as=16^2+2*32*12
v^2256+768
v=sq.rt.1024=32ft/sec
now, v=u+at
t=v-u/a=32-16/32=0.5sec
b)v=u+at
16+32*1=48ft/sec