Question

How long would it take a car with a top speed of 203 mph to travel 2.67 miles up a slope of 6.4% accelerating at 0.092 G's

Answer 1

Answer:

First you would have to multiply and divide the miles driven or travel with acceleration. Than you get 3 minutes and 41 second.

Step-by-step explanation:

Answer 2

Answer:

319.355 seconds.

Step-by-step explanation:

Ok, let us start by writting the parameters given in the question out. First is top speed,V of 203 mph which is equal to 297 feet per seconds(ft/s), the distance of 2.67 miles, the slope Percentage is 6.4 percent, that is, 6.4/100 = 0.064, and the acceleration is at 0.092 g, that is; 3 feet per seconds square(ft/s^2).

Step one: find the angle at which the slope is formed, that is;

tan θ= 0.064.

Therefore, θ= inverse of tan, tan^-1(0.064) = 3.7°.

Step two: calculate the net acceleration; this can be calculated using the formula below;

Net acceleration,a= acceleration given in the question,a(1) - (g×sin θ).

Hence, net acceleration,a= 3 - (sin 3.7° × 32).

====> a= 3- 2.1= 0.93 ft/s^2.

Step three: calculate the time taken to cover the distance. This can be calculated by using the formula below;

Top Speed,v= initial velocity, u + (acceleration,a × time taken, t).

The initial velocity is zero, therefore, v=at.

Time taken, t= v/a.

Time taken= 297/0.93.

Time taken, t= 319.355 seconds.