This question can be answered using the formula D=VT for both cyclists.

[Turkishking]

A bicylist is finishing his repair of a flat tire when a friend rides by a constant speed of 3.5 m/s. Two seconds later the bicylist hops on his bike and accelerates 2.4 m/s^2 until he catches his friend.

a) How much time does it take until he catches his friend?

What I did was:

v=v0 + at

solved for T..

V-V0/a = t

3.5 - 0/2.4 = 1.6s

(b) how far has he traveled?

Can I use D=VT? or should I use the formula x=x0 + 1/2(V0 + V)t?

I'm confused when I can use D=vt..

 

[MarkFL]

The formula:

$d=vt$

should only be applied when $v$ is constant, otherwise you should use:

$d=\overline{v}t$

where $\overline{v}$ is the average speed. When there is constant acceleration, then we know:

$\overline{v}=\frac{v_i+v_f}{2}$

 

[Turkishking]

The formula:

$d=vt$

should only be applied when $v$ is constant, otherwise you should use:

$d=\overline{v}t$

where $\overline{v}$ is the average speed. When there is constant acceleration, then we know:

$\overline{v}=\frac{v_i+v_f}{2}$

can i use x=x0 + 1/2(V0 + V)t?

 

[MarkFL]

can i use x=x0 + 1/2(V0 + V)t?

Yes, if you'll notice we can write:

$d=\overline{v}t\tag{1}$

Now, suppose we use

$d=\Delta x=x-x_0,\,\overline{v}=\frac{v_0+v}{2}$

Then (1) becomes:

$x-x_0=\frac{v_0+v}{2}t$

or:

$x=x_0+\frac{v_0+v}{2}t$

So, the two formulas are really the same thing just written in different ways.

 

[Turkishking]

Yes, if you'll notice we can write:

$d=\overline{v}t\tag{1}$

Now, suppose we use

$d=\Delta x=x-x_0,\,\overline{v}=\frac{v_0+v}{2}$

Then (1) becomes:

$x-x_0=\frac{v_0+v}{2}t$

or:

$x=x_0+\frac{v_0+v}{2}t$

So, the two formulas are really the same thing just written in different ways.

thank you :)

 

[haruspex]

3.5 - 0/2.4 = 1.6s

i presume you mean (3.5 - 0)/2.4.
That equation will tell you how long it takes for the cyclist to reach 3.5m/s, but that is not the question asked.
Your mistake is to apply equations just because they involve the right types of quantity, in this case an acceleration, a time and two speeds. An equation is meaningless divorced of the context to which it applies.

 

[Turkishking]

i presume you mean (3.5 - 0)/2.4.
That equation will tell you how long it takes for the cyclist to reach 3.5m/s, but that is not the question asked.
Your mistake is to apply equations just because they involve the right types of quantity, in this case an acceleration, a time and two speeds. An equation is meaningless divorced of the context to which it applies.

So then what am I doing wrong

 

[haruspex]

So then what am I doing wrong

I have told you what you did wrong. What you would like to know is what you should have done.
At time 0 the second cyclist passes the first. At time t, >2s, how far has each cyclist travelled?