Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h. How far must the faster car travel before it has a 15-min lead on the slower car

The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

Explanation:

Given;

speed of the faster car, v₁ = 60 mi/h

speed of the slower car, v₂ = 55 mi/h

Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles

Note: divide 15 mins by 60 to convert to hours for consistency in the units.

Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards