Question

Review Interactive Solution 2.33 in preparation for this problem. Suppose a car is traveling at +17.3 m/s, and the driver sees a traffic light turn red. After 0.226 s has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 5.00 m/s2. What is the stopping distance of the car, as measured from the point where the driver first notices the red light?

Answer 1

the distance that the car will travel before any break is applied = velocity × reaction time

l = 17.3 × .226

l =3.9098 m

now after that a deaccelerade is applied due to which it will eventually stop in the end let T be the time for it to be happen. by using newtons 1st eqn if motion

V final speed,U initial speed

V=U+aT . (a=-ve here because it is deacceleration)

0=17.3- 5×T

T= 17.3/5

T= 3.46

now using second eqn of motion in which S be the distance travelled,U initial speed,a acceleratio

S=UT+1/2(a× T^2)

S= 17.3×3.46 - 1/2(5×3.46×3.46) .(*-ve sign due to deacceleration)

S=59.858 - 29.929

S=29.929

total distance trevelled since seeing the red light= l+S

=29.929+3.9098

=33.8388 answer