Two cars 120 miles apart are driving towards each other. Car 1is driving 35 mph

Question

Two cars 120 miles apart are driving towards each other. Car 1is driving 35 mph and Car 2 is driving 45 mph. In what time (hours)will they meet? I know that one of the kinematic equations that involvevelocity initial, time, displacement, etc must be used, but I'm notsure how to set up the problem to solve for time. Two cars 120 miles apart are driving towards each other. Car 1is driving 35 mph and Car 2 is driving 45 mph. In what time (hours)will they meet? I know that one of the kinematic equations that involvevelocity initial, time, displacement, etc must be used, but I'm notsure how to set up the problem to solve for time.

Explanation / Answer

the speed of car 1 v1 = u1 + a1 * t u1 = 0 or v1 = a1 * t or t = (v1/a1) -----------(1) the speed of car 2 v2 = u2 + a2 * t u2 = 0 or v2 = a2 * t or t = (v2/a2) ------------(2) from (1) and (2) (v1/a1) = (v2/a2) or (a1/a2) = (v1/v2) let the distance covered by car 1 in time t be x therefore thedistance covered by car 2 is (120 - x) miles = (120 * 1609.344 - x) meters = (193121.28 - x) meters the distance travelled by car 1 x = u1 * t + (1/2)a1 * t^2 or x = (1/2)a1 * t^2 -----------(3) similarly,the distance travelled by car 2 (193121.28 - x) = u2 * t + (1/2)a2 * t^2 or (193121.28 - x) = (1/2)a2 * t^2 ----------(4) (3)/(4) (x/ (193121.28 - x)) = ((1/2)a1 * t^2/(1/2)a2 * t^2) =(a1/a2) = (v1/v2) = (35/45) = (7/9) or 9x = 7 *  (193121.28 - x) or 16x = 1351848.96 or x = 84490.56 m therefore the two cars meet at a distance x = 84490.56 m fromthe starting point of car 1 the acceleration of car 1 is v1^2 - u1^2 = 2a1 * x or a1 = (v1^2 - u1^2/2x) = ((35)^2 - (0)^2/2 * 84490.56) =7.25 * 10^-3 m/h^2 from (3) t = (2x/a1)^(1/2) = (2 * 84490.56/7.25 * 10^-3)^(1/2) =4827.81 h

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