Question
I really cannot understand how to get to the answer. Thank you!
Two cars are traveling, Car A is behind the other, on a straight road. Each has a speed of 21 m/s and the distance between them is 25 m. The driver of the rear car decides to overtake the Car B ahead and does so by accelerating at 2 m/s2 up to 30 m/s after which he continues at this speed until he is 25 m ahead of the other car. How far does the overtaking car travel along the road between the beginning and end of this operation?
So,
Car A:
V: 21 m/s
a: 2m/s up to V: 30m/s until it is 25 m ahead of Car B
Car B:
V: 21 m/s
a: constant = 0
I know we are looking for the total distance traveled from Car A until it overtake Car B 25 m ahead. Then I probably need to find the exact time both car are next to each other first? Then to find which position Car A is at 25m ahead.
Two cars are traveling, Car A is behind the other, on a straight road. Each has a speed of 21 m/s and the distance between them is 25 m. The driver of the rear car decides to overtake the Car B ahead and does so by accelerating at 2 m/s2 up to 30 m/s after which he continues at this speed until he is 25 m ahead of the other car. How far does the overtaking car travel along the road between the beginning and end of this operation?
So,
Car A:
V: 21 m/s
a: 2m/s up to V: 30m/s until it is 25 m ahead of Car B
Car B:
V: 21 m/s
a: constant = 0
I know we are looking for the total distance traveled from Car A until it overtake Car B 25 m ahead. Then I probably need to find the exact time both car are next to each other first? Then to find which position Car A is at 25m ahead.
Answers
scott
using car B as the frame of reference ... car A is 25 m behind
A accelerates at 2 m/s^2 up to 9 m/s ... this takes 4.5 s
... A's ave velocity during acceleration is ... 4.5 m/s
... A's distance during acceleration is ... 4.5 s * 4.5 m/s = 20.25 m
A needs to go another 29.75 m ... 50 m - 20.25 m
... this takes ... 29.75 m / 9 m/s = 3.31 s
... so the overtaking takes ... 4.5 s + 3.31 s = 7.81 s
during the overtaking , B moves ... 21 m/s * 7.81 s = 164 m
A travels 50 m more , going from 25 m behind to 25 m in front
A accelerates at 2 m/s^2 up to 9 m/s ... this takes 4.5 s
... A's ave velocity during acceleration is ... 4.5 m/s
... A's distance during acceleration is ... 4.5 s * 4.5 m/s = 20.25 m
A needs to go another 29.75 m ... 50 m - 20.25 m
... this takes ... 29.75 m / 9 m/s = 3.31 s
... so the overtaking takes ... 4.5 s + 3.31 s = 7.81 s
during the overtaking , B moves ... 21 m/s * 7.81 s = 164 m
A travels 50 m more , going from 25 m behind to 25 m in front
Darius
Just to clarity the part under the problem question is my own work to try and figure how the answer, so its not part of the problem.
So if I understand,
First we are looking for the time when the Velocity of Car B change so
Vf = Vi + ax * t
t= (Vf-Vi)/ax
t= (30-21)/2
t= 4.5 seconds
It then takes 4.5 seconds for the the Velocity of Car A to change to 30 m/s^2
But I do not get how you get a velocity of 4.5 m/s?
I do know that for the acceleration you are doing
V=(Delta x)/(Delta t)
X = (V)/(t)
also how do you get the 29.75m and the 9m/s?
So if I understand,
First we are looking for the time when the Velocity of Car B change so
Vf = Vi + ax * t
t= (Vf-Vi)/ax
t= (30-21)/2
t= 4.5 seconds
It then takes 4.5 seconds for the the Velocity of Car A to change to 30 m/s^2
But I do not get how you get a velocity of 4.5 m/s?
I do know that for the acceleration you are doing
V=(Delta x)/(Delta t)
X = (V)/(t)
also how do you get the 29.75m and the 9m/s?
scott
B is the reference frame , so look at A's motion relative to B
A accelerates to an additional 9 m/s
... the average velocity during acceleration is ... (0 + 9) / 2 = 4.5
... it covers a distance of ... 4.5 m/s * 4.5 s = 20.25 m
... it needs to go 29.75 more meters to be 25 m ahead of B
I can explain it to you ... but I can't understand it for you
A accelerates to an additional 9 m/s
... the average velocity during acceleration is ... (0 + 9) / 2 = 4.5
... it covers a distance of ... 4.5 m/s * 4.5 s = 20.25 m
... it needs to go 29.75 more meters to be 25 m ahead of B
I can explain it to you ... but I can't understand it for you