Relative velocities: braking car

[Poetria]

Homework Statement

A bicycle rider is riding at a constant speed of v(b) and at t=0 is 17m behind the car. The cyclist reaches the car when the car just comes to rest.
The car is moving with an acceleration:for 0<t<t(1)
a(c)=0

for t(1)<t<t(2)
a(c)= -c(t-t(1))

where its initial component of the velocity is v0=12m/s, t(1)=1s and c=6m/s^3. The car comes to rest at t(2).

Find the speed of the bicycle.

Homework Equations

[/B]
for t(1)<t<t(2)
v_c(t)=v_0-1/2*c*(t-t_1)^2
x_c(t)=v_0*(t)-1/6*c*(t-t_1)^3

The Attempt at a Solution

[/B]
Car: distance -
for 0<t<t(1) 12m
for t(1)<t<t(2) 4m

the time when car stops: 0=12-1/2*6*(t-1)^2 I got t(2)=3

x_c(t)=12-1/6*6*(3-1)^3

The speed of cyclist: (17+12+4)/3

Well, there is a mistake somewhere. :(

 

[scottdave]

This statement: a(c)= -c(t-t(1)) tells me that acceleration varies with time. If that is the case, then the formula x = V₀*t + (1/2)*a*t² is not valid, because that only holds for constant acceleration.

 

[scottdave]

You must integrate the expression for acceleration, to find v(t), then integrate that to find x(t). When acceleration is constant, then you get the typical equations of motion that you are familiar with.

 

[Poetria]

You must integrate the expression for acceleration, to find v(t), then integrate that to find x(t). When acceleration is constant, then you get the typical equations of motion that you are familiar with.

Ok, I will try. Many thanks.

 

[Poetria]

Integral for v(t)
v_0-1/2*c*(t-t_1)^2

Integral for x(t)
v_0*(t)-1/6*c*(t-t_1)^3

Is it correct to evaluate it as follows:
(12*3-1/6*6*(3-1)^3)-(12-1/6*6*(1-1)^3)

 

[scottdave]

If you integrate c*(t - k)dt where c and k are constants, then you get c*(t²/2 - k*t) + Integration_constant. The integration constant will equal the initial velocity, in this case. And k is equal to your t(1).

 

[Poetria]

c*(t2/2 - k*t)

and then for x I would get:
v_0*t-c*(t^3/6 - k*t^2/2)

Well, this is weird, you know. This problem has two parts and at first I thought like you, I mean I integrated c*(t_1) as c*t_1*t but this wasn't accepted. On the other hand, I can't solve the second part.

So I will use your integrals:
12-6*(t^2/2 - 1*t)=0
t_2=1+sqrt(5)

(12*(1+sqrt(5))-6*((1+sqrt(5))^3/6 - (1+sqrt(5))^2/2))-(12-6*(1^3/6 - 1^2/2))=10*sqrt(5)

Speed of the cyclist:
(17+12+10*sqrt(5))/(1+sqrt(5))

 

[haruspex]

12-6*(t^2/2 - 1*t)=0

Looks like you integrated over 0 to t₂ instead of t₁ to t₂.

 

[Poetria]

and then for x I would get:
v_0*t-c*(t^3/6 - k*t^2/2)

Well, this is weird, you know. This problem has two parts and at first I thought like you, I mean I integrated c*(t_1) as c*t_1*t but this wasn't accepted. On the other hand, I can't solve the second part.

So I will try to evaluate: v_0*t-c*(t^3/6 - k*t^2/2)

(12*3-6*(3^3/6 - 3^2/2))-(12-6*(1^3/6 - 1^2/2))=22

speed of the cyclist: 51/3

Looks like you integrated over 0 to t₂ instead of t₁ to t₂.

Looks like you integrated over 0 to t₂ instead of t₁ to t₂.

OMG, I will fix it. Many thanks.

 

[Poetria]

Is it correct:

(12-6*(t^2/2 - 1*t))-(12-6*(1^2/2 - 1*1))

t_2 = 2 ?

(12*2-6*(2^3/6 - 1*2^2/2))-(12*1-6*(1^3/6 - 1*1^2/2))

the speed of the cyclist:
(17+12+14)/2

 

[haruspex]

(12-6*(t^2/2 - 1*t))-(12-6*(1^2/2 - 1*1))

No. The v₀t term is from 0 to t₂. It's just the integral with the factor c that is from t₁ to t₂.

 

[Poetria]

No. The v₀t term is from 0 to t₂. It's just the integral with the factor c that is from t₁ to t₂.

Thanks for your patience.
The next try:

v_c(t)-v_0=-1/2*6*(t-1)^2+1/2*6*(1-1)^2
v_c(t)=12-3 (-1 + t)^2
0=12-3 (-1 + t_2)^2
t_2=3

x_c(t) - v_0*(t) =-1/6*c*(t-t_1)^3+1/6*c*(t-t_1)^3
x_c(t) - 12*3 = -1/6*6*(3-1)^3+1/6*6*(1-1)^3
x_c(t) = -8+36

(17+12+28)/3

 

[haruspex]

t_2=3

Yes.

x_c(t) - v_0*(t) =-1/6*c*(t-t_1)^3+1/6*c*(t-t_1)^3

Isn't the right hand side of that equation obviously zero? (But should not be.)

x_c(t) = -8+36

That I agree with, but I don't understand how you then get:

(17+12+28)/3

 

[Poetria]

Yes.

Isn't the right hand side of that equation obviously zero? (But should not be.)

That I agree with, but I don't understand how you then get:

Oh yes, this is for t_1<t<t_2

x_c(t) - v_0*(t) =-1/6*c*(t_2-t_1)^3+1/6*c*(t_1-t_1)^3

The bike is 17 m behind the car

for 0<t<t_1 acceleration of the car is 0
distance traveled by the car is 12m

and for t_1<t<t_2
distance traveled by the car is 28

To reach the car the bike has to travel 17+12+28 in three seconds?

 

[haruspex]

x_c(t) - v_0*(t) =-1/6*c*(t_2-t_1)^3+1/6*c*(t_1-t_1)^3

Ok, and the last term is zero, so we can simplify that to x_c(t) - v_0*(t) =-1/6*c*(t_2-t_1)^3.
But that's not quite right either; you have t on the left and t₂ on the right. So to be strictly correct it should be x_c(t) - v_0*(t) =-1/6*c*(t-t_1)^3, with t=t₂ being a value of interest.

Now, let's be clear about what x_c(t) means here. Is it only the distance the car has covered since t₁ or is it the entire distance since t=0?
The equation should also be valid at t=t₁. This gives x_c(t₁)=v₀t₁. So what exactly is x_c(t)?

 

[Poetria]

Ok, and the last term is zero, so we can simplify that to x_c(t) - v_0*(t) =-1/6*c*(t_2-t_1)^3.
But that's not quite right either; you have t on the left and t₂ on the right. So to be strictly correct it should be x_c(t) - v_0*(t) =-1/6*c*(t-t_1)^3, with t=t₂ being a value of interest.

Now, let's be clear about what x_c(t) means here. Is it only the distance the car has covered since t₁ or is it the entire distance since t=0?
The equation should also be valid at t=t₁. This gives x_c(t₁)=v₀t₁. So what exactly is x_c(t)?

Yes, t_2, it was messy.

I understand there are two stages:

0<t<t_1
x_c(t)=v_0*t

and

t_1<t<t_2

x_c(t)=v_0*(t)-1/6*c*(t-t_1)^3

I thought this was a distance from t_1 to t_2.

I see I think now I was wrong. :(

 

[Poetria]

So it would be (17+28)/3.

 

[haruspex]

So it would be (17+28)/3.

Which simplifies to...? And state the units.

 

[Poetria]

Which simplifies to...? And state the units.

15m/s

Round number.

 

[haruspex]

15m/s

Round number.

... which is always reassuring.

 

[Poetria]

... which is always reassuring.

Many thanks. :) Now I understand everything. :)