Finding the time elapsed, distance and speed

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The discussion revolves around calculating the time elapsed, distance covered, and top speed of a horse during a race with constant acceleration. The horse accelerates at 1.13 m/s² from rest, covering a total distance of 2.41 km in 2.05 minutes. The calculations suggest that the time spent accelerating is approximately 65.3 seconds, with the distance covered during this phase being around 2130 meters. The top speed reached by the horse is estimated to be 73.8 m/s, although some participants suggest this figure may be too high given the overall distance and time constraints. The conversation emphasizes the need for proper equations to relate acceleration and constant speed to accurately determine these values.
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Homework Statement


A horse race takes place on a long straight track, at the beginning of the race the horse accelerates from rest to some top speed with a (constant) acceleration of 1.13 m/s^2 and then maintains that top speed to the end of the race. The total time taken from the start of the race to its end is 2.05 minutes and the total distance covered is 2.41 km.

(a) What was the time elapsed during the acceleration phase of the horse's motion?
(b) What distance did the force cover during the acceleration phase of its motion?
(c)What was the top speed reached by the horse?

Homework Equations


None.

The Attempt at a Solution


(A)
a= 1.13 m/s^2
distance/displacement = 2410 m
t= 123s
v1= 0
v2=?

V2^2 = V1^2+2ad
V2^2 = 0+2(1.13)(2410)
v2 = 73.8 m/s


t=?
v2=73.8
v1=0
a=1.13 m/s^2

v2 = v1+at
Therefore, t = v2-v1/a

t = (73.8-0)/(1.13) = 65.3 s


(B)
t= 65.3s
a=1.13m/s^2
d= ?
V2=73.8m/s^2
v1= 0

d = v2^2-v1^2/2a
d= (73.8)-(0)/2(1.13) = 2130 m

(C) Top speed was found in (A) to be 73.8 m/s



This was my guess on how to do it...help?
 
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V2^2 = V1^2+2ad
V2^2 = 0+2(1.13)(2410)
During the acceleration phase, d is much less than 2410.
You don't know the distance, so you can't use this formula to find V2.

You could sketch a velocity graph with unknown t for the acceleration time and unknown V2. The area under the graph is the known distance, so you could write an equation for that with 2 unknowns, then use v2 = at as a second equation.

Or skip the area and write the total distance as the sum of the distance while under acceleration and the distance while at constant speed. Same two unknowns, same two equations.

Looks like it will be a quadratic equation, only one solution reasonable.
If the acceleration phase was very short, then the speed would be close to 2410/123 = 20 m/s. The final V2 will be a little larger than that to compensate for the reduced speed during the acceleration phase. Nowhere near 73, though.
 
Help!
 
geonik, do you know the two distance equations for accelerated and constant speed motion?
write the total distance as the sum of the distance while under acceleration and the distance while at constant speed
 
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