Horizontal Motion (1 Viewer)

[_ShiFTy_]

A car starts from rest to go a distance 's' in a straight line. Its acceleration 't' seconds after the start is a(1 - t/b) until it has attained maximum velocity. ('a' and 'b' are constants, and it attains this maximum velocity within a distance less than 's')

If 'T' be the whole time for the distance, and T₁ be the corresponding time for the same car to go the distance at maximum velocity, prove that

b = 3(T - T₁) and
a = 2s/3T₁(T - T₁)


So what i've done so far is this:

dv/dt = a(1 - t/b)
v = a(t - t²/2b)

dx/dt = a(t - t²/2b)
x = a(t²/2 - t³/6b)

V_max when acceleration = 0, t = b
V_max = ab/2

Not sure but does "T₁ be the corresponding time for the same car to go the distance at maximum velocity" mean:
T₁ = s/V_max
T₁ = 2s/ab?

dno where to go from here...

 

[hyparzero]

man, that question is ridiculous,

i get,

2b = 3(T - T₁)

EDIT:
====
Damn, screwed up the working out, but i finally got the correct answer: look at my next post below

 

[alcalder]

A car starts from rest to go a distance 's' in a straight line. Its acceleration 't' seconds after the start is a(1 - t/b) until it has attained maximum velocity. ('a' and 'b' are constants, and it attains this maximum velocity within a distance less than 's')

If 'T' be the whole time for the distance, and T₁ be the corresponding time for the same car to go the distance at maximum velocity, prove that

b = 3(T - T₁) and
a = 2s/3T₁(T - T₁)


So what i've done so far is this:

dv/dt = a(1 - t/b)
v = a(t - t²/2b)

dx/dt = a(t - t²/2b)
x = a(t²/2 - t³/6b)

V_max when acceleration = 0, t = b
V_max = ab/2

Not sure but does "T₁ be the corresponding time for the same car to go the distance at maximum velocity" mean:
T₁ = s/V_max
T₁ = 2s/ab?

dno where to go from here...

This is tricky. It has taken me all morning!

Right. The key words here are "acceleration 't' seconds after the start is a(1 - t/b) until it has attained maximum velocity"

So, at time t=b, the car reaches its maximum velocity and continues on at that velocity.
Maximum velocity = ab/2 m.s.
The point it reaches this maximum velocity is s = 2ab²/2 m

So, for the last bit of the car's travel from s=2ab²/2 to s=s, or the
distance = s-2ab²/2 metres
it travels with constant velocity, v = ab/2 metres/second
for t = (T-b) seconds

THUS:

s = v . t

(s- 2ab²/2 ) = ab/2 . (T-b) ---- eqn (1)

But we know that if the car travelled at it's maximum velocity for the entire distance that
s = abT₁/2

Now, doing LHS and RHS of equation (1) above.

LHS = (s- 2ab²/2 )

remove a but subbing from s = abT₁/2

Then try putting in b = 3(T-T₁)

Now do,

RHS = ab/2 . (T-b)

remove a, put in result for b.

LHS = RHS so b=3(T-T₁) [This should be a valid argument.]

determining a is just putting b back into s = abT₁/2

 

[hyparzero]

omfg, I've figured it out, without using the alcalder's LHS = RHS arguement.

Proof:
dv/dt = a(1 - t/b)
v = a(t - t²/2b)
x = a(t²/2 - t²/6b)
V_max when acceleration = 0, t = b
V_max = ab/2

Now we find the displacement when the car reaches maximum velocity, which occurs when t = b

Therefore:
x = a(b²/2 - b²/6b)
=> x = 2ab² / 6
=> x = ab² / 3 ......... (1)

T is the total time, to travel length s... therefore, the time taken to travel the rest of the way after reaching Maximum velocity is: (T - b)

But Time = Displacement/Velocity

And the displacement left after reaching V_max, is:
s - (ab² /3)

and the Velocity for that length is:
ab/2

Therefore;
(T - b) = [s - (ab² /3)] / (ab/2)
=> (T - b) = [ 2 (3s - (ab²) / 3ab ]
=> (T - b) = ( 6s - 2ab² ) / 3ab
=> (T - b) = 6s / 3ab - 2ab² / 3ab
=> (T - b) = 2s/ab - 2b/3

But T₁ = 2s/ab [Total time travelled over s at V_max]

Therefore
(T - b) = T₁ - 2b/3
=> T - T₁ = b - 2b/3
=> T - T₁ = b/3

.'. b = 3(T - T₁)

 

[alcalder]

Good one. Knew there had to be an easier way. But we both got there in the end.