Car crossing 40m intersection in 4s with 3.3 m/s² acceleration

[thomasrules]

A car is observed to cross an intersection in 4s. If the intersection is 40 meters wide, and the car accelerated at 3.3m/s^2, calculate its speed when half-way across the intersection.

I have been having a tought time with this problem because I'm not sure what the initial speed is. Also I don't know what the time would be at half. Its confusing, please help

 

[arildno]

1. You know the time when it has traveled 40meters, and the constant acceleration during that period.
Use this info to determine the initial velocity, from the equation giving you the distance traveled as a function of t.
2. Determine the value of "t" when the car has traveled 20 meters (from the same equation for the position).
Insert that value of "t" into your expression for the velocity at time "t".

 

[thomasrules]

what is the equation?

 

[thomasrules]

someone still help explain more I'm still not getting it

 

[Pyrrhus]

Use arildno reasoning with the kinematic equations for uniform acceleration

$$info:$$
$$x_{o} = 0 m$$
$$x = 40 m$$
$$a = 3.3 m/s^2$$
$$t = 4s$$


Using this equation you can get the initial speed

$$x - x_{o} = v_{o}t + \frac{1}{2}at^2$$

Now using the same equation, you can get the time t, with a displacement of 20 m and same acceleration

$$x - x_{o} = v_{o}t + \frac{1}{2}at^2$$

Now using this equation, you can plug in the time t, and find the speed at that time (at 20 m)

$$v = v_{o} + at$$

arildno answer couldn't be any more clear.