At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.50 m/s^2. at the same instant a truck, traveling with a constant speed of 9.40 m/s, overtakes and passes the automobile. (a) how far beyond the traffic signal will the automobile overtake the truck? (b) how fast will the automobile be traveling at that instant?

a)X=70.688m

b)v=18.8m/S

Explanation:

the truck has constant speed while the car moves in uniformly accelerated motion

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)

\frac{Vf^{2}-Vo^2}/{2.a} =X(2)

X=Xo+ VoT+0.5at^{2}    (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve .

For the automobile

we use the ecuation number 3

Xo=0

Vo=0

a=2.5

X=0.5(2.5)t^2

X=1.25t^2

for the truck

x=Vt

X=9.4t

the distance is the same we can match the previous equations

1.25t^2=9.4t

t=9.4/1.25=7.52s

using the ecuation for the truck

x=9.4*7.52=70.688m

b) for this point we can use the ecuation number 1 for automobile

Vf=Vo+at

Vf=0+2.5(7.52)

Vf=18.8m/S