Imagine two cars: Car A starting at 1m/s and continuing with an acceleration of 2.5m/s and Car B starting at 8m/s with no acceleration. Presuming they start from the same place, which car would win a race lasting 1 second? If the race lasted an hour, which would win? Explain your answer
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Distance traveled by A in one second is s = ut + 1/2 at^2.
1 + 2.5*1*1 = 3.5 m
Distance traveled by the other car is 8m
Car B wins
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Distance traveled by A in 60 second is s = ut + 1/2 at^2.
1 + 2.5*60*60 = 9001 m
Distance traveled by the other car is 8*60 = 480m
Car A wins
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1 + 2.5*1*1 = 3.5 m
Distance traveled by the other car is 8m
Car B wins
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Distance traveled by A in 60 second is s = ut + 1/2 at^2.
1 + 2.5*60*60 = 9001 m
Distance traveled by the other car is 8*60 = 480m
Car A wins
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U don't have to do any math to solve this one :>)
The car with the higher initial speed = B at 8 m/s will win a race of 1 s since A only has a speed of 1 + 2.5 = 3.5 m/s after 1 s.
and
the car = A with an acceleration (increasing speed) will certainly win in an hour (3600 s) over a car = B with no acceleration and a speed = 8 m/s that is surpassed in speed by car B in less than 4 s.
The car with the higher initial speed = B at 8 m/s will win a race of 1 s since A only has a speed of 1 + 2.5 = 3.5 m/s after 1 s.
and
the car = A with an acceleration (increasing speed) will certainly win in an hour (3600 s) over a car = B with no acceleration and a speed = 8 m/s that is surpassed in speed by car B in less than 4 s.
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The car with the highest d value at a given time would be the winner of a race
Givens:
ViA = 1m/s
aA = 2.5m/s^2
ViB = 8m/s
aB = 0m/s^2
The following equations will give us the distance traveled by the cars at any time, t.
dA = (ViA)t + (1/2)(aA)(t^2)
dB = (ViB)t + (1/2)(aB)(t^2) = (ViB)t
when t = 1s,
dA = (ViA)t + (1/2)(aA)(t^2)
= (1m/s)(1s) + (1/2)(2.5m/s^2)(1s)^2
= 2.25m
dB = (ViB)t
= (8m/s)(1s)
= 8m
Car B wins
when t = 1hr = 3600s
dA = (ViA)t + (1/2)(aA)(t^2)
= (1m/s)(3600s) + (1/2)(2.5m/s^2)(3600s)^2
= 1.62 x 10^7m
dB = (ViB)t
= (8m/s)(3600s)
= 2.88 x 10^4m
Car A wins
The longer the race lasts, the more favourable it becomes for Car A to win, because for every second that passes, its velocity increases by 2.5m/s. So even though it starts off slower than Car B, after just 3 seconds, it will be moving faster than Car B. A short time after that, the average velocity of Car A throughout the race to that point, will surpass the average velocity of Car B to that point. A race of that length or longer will always see Car A as the victor.
Givens:
ViA = 1m/s
aA = 2.5m/s^2
ViB = 8m/s
aB = 0m/s^2
The following equations will give us the distance traveled by the cars at any time, t.
dA = (ViA)t + (1/2)(aA)(t^2)
dB = (ViB)t + (1/2)(aB)(t^2) = (ViB)t
when t = 1s,
dA = (ViA)t + (1/2)(aA)(t^2)
= (1m/s)(1s) + (1/2)(2.5m/s^2)(1s)^2
= 2.25m
dB = (ViB)t
= (8m/s)(1s)
= 8m
Car B wins
when t = 1hr = 3600s
dA = (ViA)t + (1/2)(aA)(t^2)
= (1m/s)(3600s) + (1/2)(2.5m/s^2)(3600s)^2
= 1.62 x 10^7m
dB = (ViB)t
= (8m/s)(3600s)
= 2.88 x 10^4m
Car A wins
The longer the race lasts, the more favourable it becomes for Car A to win, because for every second that passes, its velocity increases by 2.5m/s. So even though it starts off slower than Car B, after just 3 seconds, it will be moving faster than Car B. A short time after that, the average velocity of Car A throughout the race to that point, will surpass the average velocity of Car B to that point. A race of that length or longer will always see Car A as the victor.