# Meeting point between two vehicles

• Sep 2nd 2011, 02:56 AM
roahau
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.
• Sep 2nd 2011, 03:11 AM
skeeter
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.
• Sep 2nd 2011, 03:28 AM
roahau
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
• Sep 2nd 2011, 04:18 AM
skeeter
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.