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M1 Question
A car, moving with uniform acceleration along a straight level road, passed points A and B when moving with speed 30m/s and 60m/s respectively. Find the speed of the car at the instant it passed C, the mid point of AB. The answer is 47.4m/s but I dont know how. Please help!
Reply 1
Let a be the acceleration of the car.
Let the speeds of the car at A, B and C be u, v and w respectively.
Let s be the distance between A and C (which equals the distance between C and B).
By SUVAT,
w^2 - u^2 = 2as
v^2 - w^2 = 2as
Subtracting,
2w^2 = u^2 + v^2
But we know that u = 30 and v = 60. So
w
= sqrt((30^2 + 60^2)/2)
= 47.4
Let the speeds of the car at A, B and C be u, v and w respectively.
Let s be the distance between A and C (which equals the distance between C and B).
By SUVAT,
w^2 - u^2 = 2as
v^2 - w^2 = 2as
Subtracting,
2w^2 = u^2 + v^2
But we know that u = 30 and v = 60. So
w
= sqrt((30^2 + 60^2)/2)
= 47.4
Reply 2
15
Call the acceleration a
Call the time T as it passes B
Call the distance d between A and B, though the answer should be independent of this.
Call the time t as it passes the midpoint C
Call the velocity v as it passes the midpoint C
You then get several equations.
60 = 30 + aT (from knowing the speed as you pass B)
so that aT = 30
Also you have
d = 1/2 aT^2 + 30 T (distance formula up to B)
So d = 45T (using the previous equation)
These two equations give us T and a in terms of d.
To calculate t (the time to the midpoint) we note
d/2 = 1/2 a t^2 + 30 t
Put in your expression for a in terms of d, and solve the quadratic. You now have an expression for t in terms of d. It turns out to be proportional to d.
Finally you're looking for v = 30 + at.
As your expression for a should be inversely proportional to d then the d terms cancel out in the answer which is, to be spot on, 15 sqrt(10).
If you prefer, because you've only been given speeds (ratios of distance and time) you could change your units of time (or distance) and assume T=1 (or d=1) from the off, without any loss of generality, but you might prefer just to see the d term disappear at the end.
Call the time T as it passes B
Call the distance d between A and B, though the answer should be independent of this.
Call the time t as it passes the midpoint C
Call the velocity v as it passes the midpoint C
You then get several equations.
60 = 30 + aT (from knowing the speed as you pass B)
so that aT = 30
Also you have
d = 1/2 aT^2 + 30 T (distance formula up to B)
So d = 45T (using the previous equation)
These two equations give us T and a in terms of d.
To calculate t (the time to the midpoint) we note
d/2 = 1/2 a t^2 + 30 t
Put in your expression for a in terms of d, and solve the quadratic. You now have an expression for t in terms of d. It turns out to be proportional to d.
Finally you're looking for v = 30 + at.
As your expression for a should be inversely proportional to d then the d terms cancel out in the answer which is, to be spot on, 15 sqrt(10).
If you prefer, because you've only been given speeds (ratios of distance and time) you could change your units of time (or distance) and assume T=1 (or d=1) from the off, without any loss of generality, but you might prefer just to see the d term disappear at the end.
Reply 3
13
ok c is half the distance now using V^2=u^2+2as you can have 2as
as in 60^2-30^2
that answer over 2 to as we used half the distance
then add it to 30^2 and find the square root of the answer
as in 60^2-30^2
that answer over 2 to as we used half the distance
then add it to 30^2 and find the square root of the answer
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