Question

what is the acceleration of a body moving with constant acceleration covers the distance between two points 60m apart in 60sec? its velocity as it passes the second point is 15m/sec

Answer 1

Constant acceleration implies that v(t) is a linear function am I right?

Answer 2

yes

Answer 3

So what does this tell you about your s(t)?

Answer 4

Hint: Use integration.

Answer 5

a = dv/at??

Answer 6

If a = v'(t), can we say that v(t) = u + at for some constant u?

Answer 7

Recall: Integration is the reverse of differentiation.

Answer 8

a = d^2s/dt^2??

Answer 9

Yep. But let's see it in the integral form instead. ^^
So\[v(t) = \int\limits a dt\] and \[s(t) = \int\limits v(t) dt\]

Answer 10

So you know a is a constant, so v(t) = u + at, where u is the arbitrary constant of integration right?

Answer 11

The question states that "its (the body's) velocity as it passes the second point (at t = 60) is 15m/s. So we know that v(60) = 15. Solve for u in terms of a.

Answer 12

oh.. ok ok sir im still kinda confuse but ill try to solve im not really good at integral calculus sorry

Answer 13

It's alright ^_^
v(t) = u + at
v(60) = u + a(60)
15 = u + 60a
u = 15 - 60a
v(t) = 15 - 60a + at

Answer 14

sir the V is = to 15?

Answer 15

\[s(t) = \int\limits{(15 - 60a + at)}dt = 15t + 60at + 0.5at^2 + c\]
where c = arbitrary constant of integration.

Let's use s(0) = 0
15(0) + 60(0)a + 0.5(0^2)a + c = 0
c = 0
Hence \[s(t) = 15t + 60at+0.5at^2\]
Also, s(60) = 60
15(60) + 60(60)a + 0.5(60^2)a = 60
390a = -840
a = -84/39

Answer 16

Note that v = 15 only at the point where t = 60.

Answer 17

oh.. ok so the first v you use is at rest right?

Answer 18

uhh.. sir according here the answer for a = 1.67

Answer 19

@Kira_Yamato sir?

Answer 20

I don't think so... Because you don't have the value of v(0).

Answer 21

v (final velocity) =15m^s
s (displacement) =60m
t (time taken) = 60s

acceleration =a,
initial velocity =u,

v =u+ at
15m^s = U + a(60s)
a = (15m^s -u)/60s

v^2 - u^2 =2as

(15m^s)^2 - u^2 =2*60m*(15 m^s - u)/60s)
this is what im at so far...

Answer 22

there must be choices for dis question O_O

Answer 23

its a quadratic equation, just solve for then plug in the value of u into the previous equation and solve for a.

Answer 24

So it had nothing to do with integral calculus?

Answer 25

nope..

Answer 26

this is physics lol

Answer 27

wait, is it 60 seconds or 6? ^_^ @Shin_

Answer 28

oh dang hes offline .-.

Answer 29

http://answers.yahoo.com/question/index?qid=20070908094909AAOBpps <--- similar @_@

Answer 30

http://answers.yahoo.com/question/index?qid=20090716031821AADpzfB <--- and this one too :3

Answer 31

I am such a coconut .-.

Answer 32

I have all along been using calculus to solve for kinematics that I forgot my equations of motion... :(

Answer 33

@Kira_Yamato I remember doing that years ago XD It was for science class @_@ unless this is a different subject, den I would be X_x

Answer 34

i have no idea how one would solve for this with calc lol, so maybe o.o, but yeah remember your equations for motion :P

Answer 35

@shamil98 that is what I was thinking XD

Answer 36

oh and this one :D

Answer 37

(15)^2 - u^2 =2*60*(15 - u)/60)
225 - u^2 = 30 - 2u
u^2 - 2u - 195 = 0
u = 15 or u = -13

Sub. u = 15 into a,
a = (15 - 15)/60 = 0

Sub. u = -13 into a,
a = (15 + 13)/60 = 7/15

Answer 38

@Kira_Yamato YES :D I suck XD

Answer 39

How I used it was this:

Answer 40

I don't know what that is, but great job @_@ XD

Answer 41

v^2 - u^2 =2as
plug them in now lol let the velocity(u) at the start be 0.

Answer 42

I used the internet *facepalm* >.>

Answer 43

Same here char .-. , my physics knowledge is very limited..

Answer 44

@shamil98 something we are the same on XD