Meeting point between two vehicles

Driver A starts his car and accelerates with a constant accelerasion of 0.8 m/s^2.

Driver B starts and follows on a motorcycle from the same spot, 6,0 sec later.

Driver B has got a constant acceleration of 1.8 m/s^2. (Neither of them accelerate higher than 80 km/h)

a) How long does it take driver B to catch up with driver A?

b) How long have they driven when they meet?

c) How fast do each driver drive when they meet?

I have tried the four formulas:

v = at+v0

s = 1/2 (v+v0) * t

s = 1/2 a * t^2 + v0 * t

2 * a * s = v^2 - v0^2

With no luck at all. I am confident that some calculations might be wrong.

All help is appreciated, hopefully a quick response.

Re: Meeting point between two vehicles

Quote:

Originally Posted by

**roahau** Driver A starts his car and accelerates with a constant accelerasion of 0.8 m/s^2.

Driver B starts and follows on a motorcycle from the same spot, 6,0 sec later.

Driver B has got a constant acceleration of 1.8 m/s^2. (Neither of them accelerate higher than 80 km/h)

a) How long does it take driver B to catch up with driver A?

b) How long have they driven when they meet?

c) How fast do each driver drive when they meet?

for driver A ...

$\displaystyle v_A = 0.8t$

$\displaystyle s_A = 0.4t^2$

for driver B ...

$\displaystyle v_B = 1.8(t-6)$

$\displaystyle s_B = 0.9(t-6)^2$

for (a) and (b) , set $\displaystyle s_A = s_B$ , solve for t to find the time driver A's run ... don't forget that driver B's run is $\displaystyle (t-6)$. once you get the time, find the value of $\displaystyle s_A$ and $\displaystyle s_B$ (which should be the same, right?)

for (c) , determine the values of $\displaystyle v_A$ and $\displaystyle v_B$ at the time found above.

Re: Meeting point between two vehicles

Quote:

Originally Posted by

**skeeter** for driver A ...

$\displaystyle v_A = 0.8t$

$\displaystyle s_A = 0.4t^2$

for driver B ...

$\displaystyle v_B = 1.8(t-6)$

$\displaystyle s_B = 0.9(t-6)^2$

for (a) and (b) , set $\displaystyle s_A = s_B$ , solve for t to find the time driver A's run ... don't forget that driver B's run is $\displaystyle (t-6)$. once you get the time, find the value of $\displaystyle s_A$ and $\displaystyle s_B$ (which should be the same, right?)

for (c) , determine the values of $\displaystyle v_A$ and $\displaystyle v_B$ at the time found above.

How am I supposed to determine t without the values of vA and vB ? I can't use the first formula without time, and I can't use the time forumula without the final velocity

Re: Meeting point between two vehicles

Quote:

Originally Posted by

**roahau** How am I supposed to determine t without the values of vA and vB ? I can't use the first formula without time, and I can't use the time forumula without the final velocity

when the two vehicles meet, they are at the same position.

$\displaystyle s_A = s_B$

$\displaystyle 0.4t^2 = 0.9(t-6)^2$

solve for t , then proceed as I stated in my previous post.