Question

A Porsche challenges a Honda to a 400-m race. Because the Porsche's acceleration of 3.5 m/s² is larger than the Honda's 3.0 m/s², the Honda gets a 1.0 s head start. Who wins, and by how much time?

Answer 1

Answer: The Porsche Wins. Arrives 19secs earlier before the Honda.

Explanation: The head start of 1secs corresponded with the difference in acceleration {3.5m/s² - 3m/s²}=0.5m/s²

Using the first equation of motion we obtain the velocity which correspond to this acceleration (0.5m/s²) and time of 1secs where initial velocity u = 0

V = u + at

V= 0 + at

V = 0.5 * 1 = 0.5m/s

Now let find the velocity of each of the car.

If V. a

0.5m/s 0.5m/s²

........m/s. 3.5m/s²

Velocity of porche Vp

= (3.5/0.5) * 0.5 = 3.5 m/s

Also if. V. a

0.5m/s. 0.5m/s²

.....m/s. 3 m/s²

Velocity of Honda Vh

= {3/0.5} * 0.5 = 3m/s

So let's find the time t taken by both cars to cover the distance of 400m

Recall,

Velocity = distance/time

Time t = distance/velocity

For Porche, t = 400/3.5 =114.29secs

For Honda, t = 400/3 = 133.33secs

Looking critically, we noticed that Porche car took shorter time.

The difference In time is

= (133.33 - 114.29)secs = 19.04secs