Chevy word problem

[dvo314]

This problem is actually from a new Chevrolet commercial and I'm having trouble setting it up. Im familiar with d=r(t) problems but this one is much more difficult for me. The problem is:

A car is traveling at 52mph. Another car, 170.2 feet in front of it, is traveling in the same direction at 27mph. How long until they collide?

 

[Deleted member 4993]

This problem is actually from a new Chevrolet commercial and I'm having trouble setting it up. Im familiar with d=r(t) problems but this one is much more difficult for me. The problem is:

A car is traveling at 52mph. Another car, 170.2 feet in front of it, is traveling in the same direction at 27mph. How long until they collide?

Relative to the front car, the rear car is travelling at (52-27=) 25 mph. It has to travel170.2 ft.

Continue...

 

[dvo314]

Alright, so with d/r=t. That wouldd be 170.2'/25mph = t. After converting mph to feet per mile, I have 170.2/132,000=.001 hours or 3.6 seconds. Did I do that correctly?

 

[ksdhart]

You can always check your answer by plugging it in to the original equation. For simplicity's sake let's say that car 1 starts at 0 feet, and let the variable t​ represent the number of seconds elapsed. The two vehicles will collide when:

0 + t * (Car 1's speed in feet/sec) = 170.2 + t * (Car 2's speed in feet/sec)

So, then what is car 1's speed? And what is car 2's speed? Then when t = 3.6 (your solution), are the two sides equal?

 

[Dale10101]

Also

This problem is actually from a new Chevrolet commercial and I'm having trouble setting it up. Im familiar with d=r(t) problems but this one is much more difficult for me. The problem is:

A car is traveling at 52mph. Another car, 170.2 feet in front of it, is traveling in the same direction at 27mph. How long until they collide?

Final position = initial position + distance traveled = initial position + rate x time

1mph = (1m/hr)(5280 ft/m)(hr/3600s) = 1.467 ft/sec

52mph = (52m/hr)(1.467 ft/s) = 76.3 ft/s

27mph = (27m/hr)(1.467 ft/s) = 39.58 ft/s

Car 1's position = 0 + (76.3)(t)

Car 2's position = 170.2 + (39.58)(t)

Collision when Car 1's position = Car 2's position, solve for t

or, more cleverly

25mph = 25(1.467 ft/sec) = 36.675 ft/s

170.2 ft/(36.675 ft/s) = t