Calculate Car Speed Across Intersection | Challenging Question

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To calculate the car's speed halfway across a 40-meter intersection, the initial speed must first be determined using the kinematic equation for distance with constant acceleration. Given the car accelerates at 3.3 m/s² and takes 4 seconds to cross, the initial speed can be calculated. Next, the time to reach 20 meters can be found using the same distance equation, allowing for the calculation of speed at that midpoint. By substituting the time into the velocity equation, the final speed at 20 meters can be derived. This method effectively utilizes kinematic equations for uniform acceleration to solve the problem.
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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
 
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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".
 
what is the equation?
 
someone still help explain more I'm still not getting it
 
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.
 
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