A car travels along a straight road, heading east for 1 h, then traveling for 30 min on another road that leads northeast. If the car has maintained a constant speed of 52 mi/h, how far is it from its starting position

The car is 72.75 miles away from its starting position.

Explanation:

First, remember the relation:

distance = time*speed.

Also, the distance between two points (a, b) and (c, d) is:

D = √( (a - c)^2 + (b - d)^2)

For this problem, we can assume:

The North is equivalent to the y-axis, and the East is equivalent to the x-axis.

We also assume that the initial position of the car is (0mi, 0mi)

Now the car moves to the East at a speed of 52mi/h for one hour, then the new position of the car is:

(0mi, 0mi) + (52mi/h*1h, 0mi) = (52mi, 0mi)

Now the car travels 30 mins (or 0.5 hours) to the northeast at a speed of 52mi/h.

We can assume that it moves at an exact angle of 45° from East to North, then the components of the speed can be written as:

Sx = speed in the x-axis = 52mi/h*cos(45°) = 36.77 mi/h

Sy = speed in the y-axis = 52mi/h*sin(45°) =  36.77 mi/h

Then the new position of the car is:

(52mi, 0mi) + (36.77 mi/h*0.5h, 36.77 mi/h*0.5h) = (70.385 mi, 18.385 mi)

Now we know the final position of the car.

The distance between the final position (70.385 mi, 18.385 mi) and the initial position (0mi, 0mi) is:

D = √( (70.385 mi - 0mi)^2 + (18.385 mi - 0mi)^2) = 72.75 mi

The car is 72.75 miles away from its starting position.

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explanation:

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s