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 3.7 m/s2 for 4.5 seconds. It then continues at a constant speed for 8.6 seconds, before applying the brakes such that the carâs speed decreases uniformly coming to rest 229.73 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.(a) How fast is the blue car going 3.6 seconds after it starts? (b) How fast is the blue car going 8.8 seconds after it starts? (c) How far does the blue car travel before its brakes are applied to slow down?

Answer 1

Answer:(a)13.32m/s

(b)16.65m/s

(c)180.65 m

Explanation:

Given

blue car starts from rest and accelerate for 4.5 sec at 3.7 $m/s^2$

then runs at a constant velocity  for 8.6 sec and finally decelerate to come at rest.

total distance(s)=229.73

(a)Velocity of blue car after 3.6 sec

v=u+at

here u(initial velocity )=0

v=0+(3.7)3.6=13.32 m/s

(b)Given car accelerate for 4.5 sec and after that it runs at constant velocity therefore velocity of car after 8.8 sec will equal to velocity of car after 8.6 sec

v=u+at

v=0+(3.7)4.5=16.65 m/s

(c)Distance traveled by blue car before brake is applied

let $S_1$ be the distance traveled during acceleration of car and $s_2$ be the distance traveled during constant velocity

$S_1=ut+\frac{1}{2}at^2$

here u=0

$S_1=\farc{1}{2}(3.7)(4.5)^2$=37.4625m

$S_2=vt+\frac{1}{2}at^2$

here v=16.65 m/s a=0

$S_2$=16.65(8.6)=143.19 m

Thus total distance=$S_1+S_2$=180.6525 m