Acceleration as a function of time.?

Question

Okay i have two questions here actually. I tried figuring out the whole night but to no avail. So if you guys could help, it'll be a great thing.

Q1: A particle moves along a straight line and t seconds after passing through an origin O the velocity of the body is given by v=kt^2 - ct, where k and c are constants. When t=2 the body is again at O and has an acceleration of 6 ms^-2. Find the values of k and c.

Q2: A body moves along a straight line with acceleration given by a=7t/36, where t is the time in seconds. When t=0 the body is at rest at an origin O. The acceleration continues until t=6, whereupon it ceases and the body is retarded to rest. During this retardation the acceleration is given by a=-t/4. Find the value of t when the body comes to rest and the displacement of the body from O at that time.

pls pls pls guys. somebody. i'm desperate. hehe =p. thanks yah.

Answer 1

Okay, first question. At t=0, the particle's at O; at time t>0, it has a velocity v = kt^2 - ct; and at t=2, the particle's back at O with an acceleration of 6 m/s^s.

With that velocity, the acceleration is not constant. So we take the derivative:

v(t) = kt^2 - ct
a(t) = dv/dt = 2kt - c
a(2) = 6 = 4k - c (Eq 1)

That takes care of the acceleration. Now for the displacement, we integrate v(t):

s(t) = (1/3)kt^3 - (1/2)ct^2 + so, where so, the initial displacement, is zero. Now find s(2):

s(2) = 0 = (8/3)k - 2c (Eq 2)

Solve equations (1) and (2) for k and c.

Now for the second question. Looks like there's two parts to this one. Let's do the first part first.

a(t) = 7t/36
v(t) = (7/72)t^2 (initial velocity vo is zero)
s(t) = (7/216)t^3 (initial displacement is zero)

At t=6, we have
v(6) = 7/2
s(6) = 7

Now for the second half. For convenience, let's start it over at t=0 with an initial velocity v(0) = vo = 7/2. The equations are:

a(t) = -t/4
v(t) = (-1/8)t^2 + vo = (-1/8)t^2 + 7/2
s(t) = (-1/24)t^3 + (7/2)t (initial displacement zero)

It comes to rest when v(t) = (-1/8)t^2 + 7/2 = 0
t^2 = 28 ==> t = sqrt(28) = 2 sqrt(7) (Eq 3)

Now you have one answer. When the body comes to rest,
t = 6 + 2 sqrt(7)

The displacement during the deceleration phase is

s(2 sqrt 7) = (-1/24)(2 sqrt 7)^3 + (7/2)(2 sqrt 7)
(2 sqrt 7)^3 = 56 sqrt 7
s = (-56/24) sqrt(7) + 7 sqrt(7) = (112/24) sqrt(7)
s = (14/3) sqrt(7)

To get the total displacement, you have to add the displacement from phase one. The total displacement is

s = 7 + (14/3) sqrt(7) = (35/3) sqrt(7)

That's your other answer.

Hope I didn't make a mistake.

[ed. note] I just noticed Scott's post. He solved for k and t, so I just did too. I get c=3 and k = 9/4. That doesn't match Scott's result; it appears he missed a sign change ... (-12) should be (+12). Of course, I might've missed something too. Be sure to check my work.

Answer 2

Q1: The particle ends up at the same place it started at t=0. Therefore the total distance traveled is 0. The distance is given by

integral(0 to 2) of v(t)*dt, or Integral (0 to 2) of kt^2-ct. The integral is (k/3)*t^3 - (c/2)*t^2. Evaluated at t = 2 this is (8*k)/3 - (c/2)*4 or

(8/3)k=2c; this is the first relation between c and k. The second is given by the acceleration at point t=2, given as 6*10^6 sec^-2. The acceleration is d(v)/dt, or 2kt - c. At t=2 this gives the second relation 4k-c=6*10^6. or k=c/4 +1.5*10^6

From the first relation c = 4k/3; from the second k = k/3 + 1.5*10^6,
or 2k/3=1.5*10^6, or k=2.25*10^6. again from the firs, c = 4k/3 or c=(10/3)*10^6


Q2:

The velocity is (7/36)(t^2)/2+K. Since v=0 at t=0, K=0

The distance covered in that time is integral v dt or (7/36)(t^3)/6. At t=6 this is (7/3).

In the second part, the deceleration is t/4, so velocity is -(t^2)/8+K. At t=6 in this leg the velocity is the same as the end of the first leg, which is 7/2. So v(6) = -(36/8) + K =7/2. So K = 7/2 +36/8 or 8
This gives v(t) for the second leg = 8-(t^2)/8. The velocity is zero when the body comes to rest so the time is given by

(t^2)/8 = 8 or t = 8

The displacement for the second half is Integral (6 to 8) v(t)
or 8t - (t^3)/24 [6 - 8] 64 - 512/24 - 48 + 216/24 = 11/3.

The total displacement is then 7/3+11/3 = 6

Check my numbers in case of an arithmetic error.

Your statement did say the acceleration for Q1 was 6 ms^-2 which I take to mean milliseconds ^-2, which should be 6*10^6 sec^-2

Answer 3

costly 13 12 months dude , do exactly not forget approximately velocity carry out = mandatory of acceleration perform function perform = elementary of velocity carry out a) a(t) = 15?t ? 2/?t v(t) = 10 t^(3/2) ? 4?t + C -----> v(a million) = 4 ----> C = ?2 v(t) = 10 t^(3/2) ? 4?t ? 2 b) s(t) = 4 t^(5/2) ? 8/3 t^(3/2) ? 2t + C -----> s(a million) = 0 ----> C = 2/3 s(t) = 4 t^(5/2) ? 8/3 t^(3/2) ? 2t + 2/3 s(sixteen) = 4 (sixteen)^(5/2) ? 8/3 (sixteen)^(3/2) ? 2(sixteen) + 2/3 = 3894

Answer 4

v=kt^2-ct. dv/dt=a= 2kt-c given a =6 and t=2 6=4k-c
integrate the v for s=1/3kt^3-1/2ct^2. s=0 and t=2 we have
0=8/3k-2c. Solve eq 1 for c=4k-6 and plug into the formula.
8/3k-2(4k-6)=0
8/3k-8k-12=0
8/3k-24/3k=12
-18/3k=12
k=-36/18
k=-2
use c=4(2)-6
c=2

use the same method for the other problem. If you need help email me and I will help you.

Answer 5

good answer above i just did it and was about to post it when i refreshed my browser and saw you got the same answer, but you beat me to it... it is the correct method to use...

Answer 6

zero

Answer 7

a=-t/4>>k=t*ct....