At the instant the traffic light turns green, an automoblile that has been waiting at an intersection starts ahead with a constant acceleration of 2.50 m/s^2. At the same instant, a truck traveling with a constant speed of 15.0 m/s overtakes and passes the automobile. How far beyond the starting point has the automobile traveled when it overtakes the truck? What is the automobiles speed at that point?

Question

loser66 · Answer

@ybarrap  please.

anonymous · Answer

Knowing the initial velocity and acceleration of the first and second car, you can use the distance formula (distance = initial velocity + 1/2 acceleration times time squared) to find the time. From there, you can use the distance formula properly to find the distance. And from there, acceleration by using the final velocity formula (final velocity squared = initial velocity squared + two times the acceleration times the distance). Hope that helps! And makes sense.

anonymous · Answer

thanks :p

ybarrap · Answer

Let $v_T$ be speed of the truck, $t$, the time the car passes the truck and $a_A$, the acceleration of the car.

The distance traveled by the Truck in time $t$ is $v_Tt$. The distance traveled by the Automobile in time $t$ is $1/2a_At^2$. When the automobile catches up, these will be equivalent. Solve for $t$ to find when they meet and then plug back into  $v_Tt$ to find distance traveled when they do meet.

The speed of the Automobile will be $v_A=a_At$ at the instant they meet up.

Let me know if you have any questions.

anonymous · Answer

For the truck, the distance is $$d_T=v_T·t$$For the car the distance is:$$d_C=\frac{ 1 }{ 2 }a_Ct^2$$Both distances are the same when the car reaches the truck$$d_T=d_C \rightarrow v_Tt=\frac{ 1 }{ 2 }a_Ct^2 \rightarrow t=\frac{ 2v_T }{ a_C }$$
1.-Distance by the car:$$d_C=\frac{ 1 }{ 2 }a_C \left( \frac{ 2v_T }{ a_C } \right)^2=\frac{ 2v_T^2 }{ a_C }=\frac{ 2·15^2 }{ 2.5 }$$2.-Car speed$$v_C=a_C·t=a_C \frac{ 2v_T }{ a_C }=2v_T=2·15$$