In: Physics
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 3.6 m/s2 for 4.4 seconds. It then continues at a constant speed for 8.2 seconds, before applying the brakes such that the car
part a
From ist equation of motion
Vf =Vi+at
Here Vi=0
Vf =at = 1.8(3.6)
Vf = 6.48m/s
b) we know that the velocity after 4sec will be
V =at =3.6(4.4) = 15.84m/s
After 4sec time the blue car moves 8.2sec more with constant speed. This simply means the speed of the car after 9.8sec will be 15.84m/s
c) Distance cover during ist 4.4sec
2as1 =Vf2 -Vi2
s1= Vf2 -VI2 / 2a = 14.42 -02 / 2(3.6) = 28.8m
Distance covered during next 8.2sec
S2=Vt =15.84(8.2) = 129.88m
Adding the distances
S =s1+s2
=28.8+129.88
=158.68m
d) a=?
Distance cover during deacceleration is 208.3-158.68 = 49.61m
From third equation of motion
a = Vf2 -VI2 / 2S ( In this case Vf=0m/s, Vi =15.84m/s)
a = -15.842 /2(49.61) = -2.55m/s2
e) Time taken during deacceleration is
Vf =Vi+at
t =- Vi/a =- 15.84/ -2.55 = 6.2sec
Total time = 4.4+8.2+6.2
= 18.8sec
f ) The acceleration of yellow car will be
s=vit+0.5at2 = 0+ 0.5at2
a = s/o.5t2 = 208.3 / 0.5(18.8) = 22.15m/s2