2 motion questions
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having trouble with these also :l
1)
A car starts from rest with an initial acceleration 6.5 m/s^2. The acceleration decreases linearly with time to zero in 11 s. After this, the car travels at constant speed. Determine the time required for the car to travel 542 m from the starting position.
2)
The velocity of a particle moving along the x axis is given by v = 0.1498x, where v is in mm/s and x is in mm.
If x=7.8mm when t=0, determine the value of acceleration when t=9.6 s.past questions:
a)
Given a = 8v/100
where a is in m/s^2 and v is in m/s
and that when t= 2s, v = 5 m/s;
what is the velocity by the time t = 10s?For question a, it seems i cant integrate normally..
b)
A rocket is fired vertically upward from the surface of the earth. the thrusters are controlled so that the rocket has a constant upward acceleration of 5m/s^2. The thrusters are switched off after 23s from launch. Find the maximum height it reaches before starting to fall bak to the earth.
after the thrusters it will still have an upward velocity.. -
k i try ah..
a)
a = 8v/100
a = dv/dt
8v/100 = dv/dt
so by variable separable,
1/8v dv = 1/100 dt
integrate both sides,
1/8 ln v = t/100 + C
ln v = 8t/100 +C1, where C1 = 8C
take exp of both side,
v = [e^(8t/100)]*e^C1
when t=2, v=5, so sub in,
5 = [e^0.16]*[e*C1]
den can find e^C1, den sub but in to the eqn since its a constant,
den sub t=10 in..
v = [e^0.8]*[e^C1]
u should get v = 9.48m/s
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for b) u must split it into two segments
the 1st segment u must find the velocity of the rocket at the point the thrusters were turned off.
to do this, use v= u +at
v = 0 + 5*23
v= 115m/s
so at the point when thrusters were turned off, the rocket got velocity of 115m/s
so to find its max height reached, use v^2 = u^2 +2as
v here is the velocity when it reaches its max height, which is basically zero..
so its 0 = 115^2 + 2*9.81*s
from there find s..
but this 's' is only the distance it travelled AFTER the thrusters switched off..
so to find the distance travelled before turned off, i think u can do la hor.. cannot spoonfeed.. s = ut + 1/2*a*t^2..
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ok the qn 1 i buay hiao.. so i skip.. haha.. but the qn 2.. its the same as qn A, but a few more steps..
so 1st u form the equation..
v = 0.1498x,
v = dx/dt,
so 0.1498x = dx/dt
by variable separable,
1/(0.1498x) dx = 1 dt
integrate, u get
(1/0.1498) ln x = t + C
sub in t = 0, x = 7.8, u can find C, den sub value of C into eqn, and den sub t = 9.6s, u will get x1 = 32.86
now u have the distance travelled, which is x1-x = 32.86-7.8 =25.05mm
and u have the initial velocity, which is u= 0.1498(7.8) and final velocity v = 0.1498(32.86) so by using v= u +at, where t = 9.6s, u shud find acceleration.. agn, no spoonfeeding.. lol..
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thanks.. later i will check. got to finish my essay off. will post after dinner : D
your part a) is correct : D and part b) : D -
ok haha.. whats the ans..? i can work better with answer haha.. or mods can help when they come online bah haha..
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i dun know : O
ok for q2)
i got u = 1.16844
v = 4.802588
t = 9.6s
using v = u + at
4.802588 = 1.16844 + a(9.6)
a = 0.378557 m/s^2
but it's incorrect. anythign wrong? -
v= 0.1498(32.86) = 4.922428, not 0.1498(32.06)... ur answer of 4.802588 came from the latter..
so v = u + at should be
4.922428 = 1.16844 + a(9.6)
a = 0.391m/s^2
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2)
The velocity of a particle moving along the x axis is given by v = 0.1498x, where v is in mm/s and x is in mm.
If x=7.8mm when t=0, determine the value of acceleration when t=9.6 s.as purple dragon has done
and you get C = 13.7
then you need to go back to the equation
a = dv/dt = dv/dx * dx/dt, where dv/dx = 0.1498
at t = 9.6, dx/dt = v = 0.1498 * 32.8 = 4.913thus, a = 0.736 m/s^2
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Originally posted by purpledragon84:
v= 0.1498(32.86) = 4.922428, not 0.1498(32.06)... ur answer of 4.802588 came from the latter..
so v = u + at should be
4.922428 = 1.16844 + a(9.6)
a = 0.391m/s^2
your kinematics equation is for constant acceleration. In our case here, the acceleration changes constantly.
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1)
A car starts from rest with an initial acceleration 6.5 m/s^2. The acceleration decreases linearly with time to zero in 11 s. After this, the car travels at constant speed. Determine the time required for the car to travel 542 m from the starting position.at t = 0, a = 6.5
at t = 11, a = 0from there, we can see that a = 6.5 - 6.5t/11
a = dv/dt = 6.5 - 6.5t/11
so v = 6.5t - 6.5t^2/22 + cat t = 0, v = 0, so c = 0
thus, dx/dt = v = 6.5t - 6.5t^2/22
x = 3.25t^2 - 6.5t^3/66 + C
t = 0, x = 0, so C = 0so at t = 11, x = 292.42m
v = 35.75m/sremaining distance = 542-292.42 = 249.58m
total time taken = 249.58 / 35.75 + 11 =17.98 s
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oh yah ah.. coz the qn nvr state changing acceleration la.. lol..
but wait...
"at t = 0, a = 6.5
at t = 11, a = 0from there, we can see that a = 6.5 - 6.5t/11"
but i cannot see.. lol.. how u get..?
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Originally posted by purpledragon84:
oh yah ah.. coz the qn nvr state changing acceleration la.. lol..
but wait...
"at t = 0, a = 6.5
at t = 11, a = 0from there, we can see that a = 6.5 - 6.5t/11"
but i cannot see.. lol.. how u get..?
qn nvr state, but can integrate to see mah
Anyway that method using calculus foolproof also... acc whether constant or not also can useAnd that one ah... simple sec 2... erm... find gradient and y-intercept lor...
Just that TS now uni level, this type of thing should be able to see and do mentally, so I never do out lor... -
Originally posted by eagle:
qn nvr state, but can integrate to see mah
Anyway that method using calculus foolproof also... acc whether
constant or not also can useAnd that one ah... simple sec 2... erm... find gradient and y-intercept lor...
Just that TS now uni level, this type of thing should be able to see and do mentally, so I never do out lor...
oh yah hor.. i wasn't thinking along that line... LOL.. i wanted to integrate, but couldn't form eqn.. lol.. -
Originally posted by eagle:
so at t = 11, x = 292.42m
\
v = 35.75m/sis that a mistake? x = 262.1666 right ?
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yup my mistake there... But you get the drift I hope...
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yup ^^ thanks for the help guys