Question

Luna and Angela are competitors in a running competition. Both are leading the race and enter at the same time the last 30 meters. Angela enters the final segment runnning at 4 m/s and "applies" an acceleration of 2 m/s2 while Luna enters the final segment at 3m/s but "applies " an acceleration of 3 m/s2. (a) Who finishes the race first? (Do the calculations) (b) Who is running faster at the end of the race?

Answer 1

(a) Luna

(b) Luna

Step-by-step explanation:

(b)

The final speed can be found by using the third equation of motion:

`2as = v_f^2 - v_i^2\\`

where,

a = acceleration

s = distance covered

vf = final speed

vi = initial speed

For Angela:

a = 2 m/s²

s = 30 m

vi = 4 m/s

vf = v_fa

therefore,

`2(2\ m/s^2)(30\ m) = (v_(fa))^2 - (4\ m/s)^2\\v_(fa)^2 = 120 m^2/s^2 + 16\ m^2/s^2\\v_(fa) = 11.66\ m/s\\`

For Luna:

a = 3 m/s²

s = 30 m

vi = 3 m/s

vf = v_fl

therefore,

`2(3\ m/s^2)(30\ m) = (v_(fa))^2 - (3\ m/s)^2\\v_(fa)^2 = 180 m^2/s^2 + 9\ m^2/s^2\\v_(fa) = 13.74\ m/s\\`

Therefore, Luna will be running faster at the end of the race.

(b)

To calculate the time of completion we will use the first equation of motion:

`v_f = v_i + at`

FOR ANGELA:

`11.66\ m/s = 4\ m/s + (2\ m/s^2)t_a\\t_a = (7.66\ m/s)/(2\ m/s^2)\\\\t_a = 3.83\ s`

FOR LUNA:

`13.74\ m/s = 3\ m/s + (3\ m/s^2)t_l\\t_l = (10.74\ m/s)/(3\ m/s^2)\\\\t_l = 3.58\ s`

Hence, LUNA finishes the race first.