Question

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 5.6 m/s² for 3 seconds. It then continues at a constant speed for 7.8 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 178.16 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.

1)

How fast is the blue car going 2.4 seconds after it starts?

2)

How fast is the blue car going 7.8 seconds after it starts?

3)

How far does the blue car travel before its brakes are applied to slow down?

4)

What is the acceleration of the blue car once the brakes are applied?

5)

What is the total time the blue car is moving?

6)

What is the acceleration of the yellow car?

Answer 1

here,

accelration of blue car , ab = 5.6 m/s^2

1)

after t = 2.4 s

the speed after 2.4 s , v = u + a * t

v = 0 + 5.6 * 2.4

v = 13.44 m/s

the speed of blue car is 13.44 m/s

2)

afetr t = 7.8 s

as the car is moving with the constant speed for next 7.8 s

then the speed of car is 13.44 m/s

3)

distance travelled by blue car before it deaccelrates , s = 0.5 * a * 3^2 + v * 7.8

s = 0.5 * 5.6 * 3^2 + 13.44 * 7.8

s = 130.03 m

4)

distance when car is deacccelrating , x = 178.16 - 130.03

x = 48.13 m

let the accelration be a

v^2 - u^2 = 2 * a * s

0 - 13.44^2 = 2 * a * 48.13

a = 1.88 m/s^2

the acceleration of the blue car once the brakes are applied is 1.88 m/s^2