posted by Hayls on .
Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3.5 m/s2 for 3.6 seconds. It then continues at a constant speed for 10.1 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 167 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.
How far does the blue car travel before its brakes are applied to slow down?
Add the distance travelled by the blue car while accelerating for 3.6 s to the distance travelled at constant speed for 10.1 s. You don't need the information about the yellow car or the stopping distance.
okay well I got stuck, my new question is how do I find the distance while the car is accelerating at the 3.5 m/s^2 for 3.6 seconds?
that is easy actually.....
d(distance)= 1/2at^2 +v(initial)t, where a-acceleration, t-time. v-initial here is 0 because the car starts from rest....so d= 1/2at^2 which is .5(3.5)(3.6^2)...which comes out to 22.68m I believe