- #1
Gadget
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- 0
Hello everyone! I just registered to this board, and need some help solving a physics problem. It reads:
Car A leaves a city and travels along a straight road for 1.5 min at 60km/h. It then accelerates uniformly for 0.25 min until it reaches a speed of 80 km/h. It proceedes at that speed for 2.0 min, then decelerates uniformly for 0.50min until it comes to rest. Car B leaves the same city along the same road and accelerates uniformly for 1.6 min until it reaches a speed of 160km/h. It then decelerates uniformly until it comes to rest again after 1.6 min. How far will the two cars have traveled during the different stages?
This is as far as I completely understand it:
Car A:
1 step: conversion 60km/h = 1 km/min
therefore = 1.5 km
2 step: conversion 80km/h = 1.33 km/min
therefore v = vo + at
1.33 - 1 = a = 1.32km/min
--------
.25
here is confusion:
x = x0 + v0t + 1/2at^2
x = 1.5 + 1.33(.25 min) + (1/2)(1.32)(.25)^2
or
x = (1/2)(1.32)(.25)^2
?
Car A leaves a city and travels along a straight road for 1.5 min at 60km/h. It then accelerates uniformly for 0.25 min until it reaches a speed of 80 km/h. It proceedes at that speed for 2.0 min, then decelerates uniformly for 0.50min until it comes to rest. Car B leaves the same city along the same road and accelerates uniformly for 1.6 min until it reaches a speed of 160km/h. It then decelerates uniformly until it comes to rest again after 1.6 min. How far will the two cars have traveled during the different stages?
This is as far as I completely understand it:
Car A:
1 step: conversion 60km/h = 1 km/min
therefore = 1.5 km
2 step: conversion 80km/h = 1.33 km/min
therefore v = vo + at
1.33 - 1 = a = 1.32km/min
--------
.25
here is confusion:
x = x0 + v0t + 1/2at^2
x = 1.5 + 1.33(.25 min) + (1/2)(1.32)(.25)^2
or
x = (1/2)(1.32)(.25)^2
?