Question

A truck is at a stoplight. When the light turns green, it accelerates at 2.5m/s^2. At the same instant, a car passes the truck going 15m/s. Where and when does the truck catch up with the car?

Answer 1

\[a*t^{2}/2=v*t\]
t=15*2/2.5=12 seconds

Answer 2

how do you get a * t^2 /2 ?

Answer 3

its a common equation of motion

Answer 4

\[distance=\frac12at^2+v_ot\]
when does distance equal distance?
\[\frac12a_1t^2+v_{1o}t=\frac12a_2t^2+v_{2o}t\]
assuming the car is traveiling at a constant speed, its accelration is zero, and since the truck starts from a stop, its initial velocity is zero
\[\frac12a_1t^2=v_{2o}t\]
\[\frac12a_1t^2-v_{2o}t=0\]
if we need to, we can use the quadratic equation, but that does seem to be a bit of overkill .... but i am a math major soooo;
\[t=\frac{v_{2o}\pm\sqrt{(v_{2o})^2-4(\frac12a_1)}}{2(\frac12a_1)}\]

Answer 5

\[\frac12a_1t^2=v_{2o}t\]
\[\frac12a_1t=v_{2o}\]
\[t=\frac{2v_{2o}}{a_1}\]