Question

A Porsche challenges a Honda to a 400 m race. Because the Porsche's acceleration of 3.4 m/s2 is larger than the Honda's 3.0 m/s2, the Honda gets a 100 m head start - it is only 300 m from the finish line. Assume, somewhat unrealistically, that both cars can maintain these accelerations the entire distance.

Answer 1

Answer:

Winner wins by 0.969 s

Explanation:

For the Porche:

Given:

Displacement of Porsche s = 400 m

Acceleration of Porsche a = 3.4 m/s^2

From Newton's second equation of motion,

$s = ut + (1/2) a t^2$ (u = 0 as the car was initially at rest)

Substituting the values into the equation, we have

$t^2 = (2 * 400) / 3.4$

= 235.29 / 3.4

t = 15.33 s

For the Honda:

Displacement of Honda = 310 m

Acceleration of Honda = 3 m/s^2

Applying Newton's second equation of motion

$s = ut + (1/2) a t^2$ (u = 0 for same reason)

Substituting the values into the equation, we obtain

$t^2 = (2 * 310) / 3$

= 620 / 3

t = 14.37 s

Hence

The winner (honda) wins by a time interval of = 15.33 - 14.37    

=0.969 s