starting from 1.5 miles away a car drives towards a speed check point and then passes it.

Question

can you help me solve this? starting from 1.5 miles away a car drives towards a speed check point and then passes it.The car travels at a constant rate of 53 miles per hour. The distance of the car from the checkpoint is given by d=l1.5-53tl. At what times is the car 0.1 miles away from the check point?

Answer 1

The distance from the car to the speed check point at any time is given by:

d = abs( 1.5 - 53t )

The distance from the car to the check point is 1.5 miles

solving the absolute value function with d=1.5 miles will give the time from where the car starts to the check point

there are two solutions d = 1.5 -53t , and d = - ( 1.5 -53t) for the absolute value function

1.5 = 1.5 - 53 t, 1.5 -1.5 = 53 t , t =0

1.5 = - 1.5 + 53t , 3.0 = 53t, t = 3/53 = 0.0566 hrs ( 3600 sec/hr) = 203.8 sec = 3.40 min

This is only the time to the check point, but the car travels an additional 0.1 miles past the check point. Knowing that the car was still cruising at constant speed, the time is simply found as :

t = x / v , t = ( 0.1 miles / 53 miles/hr) = 0.001886 hrs ( 3600 sec /hr) = 6.8 sec = 0.11 min

the total time from when the car starts to the point 0.1 miles away or a total distance of 1.6 miles is:

t _tot = 203.8 sec + 6.8 sec = 210. 6 sec call is 211 sec , or 3.40 min + 0.11 min = 3.51 min