In 1967, new zealander burt munro set the world record for an indian motorcycle, on the bonneville salt flats in utah, with a ma
ximum speed of 82.1 m/s. the one-way course was 8.045 km long. if it took burt 4.00 s to reach a velocity of 26.82 m/sec, how long (in seconds) did it take burt to reach his maximum speed? how far (in meters) did he travel during his acceleration? look at the equations in the last section. find acceleration first, then the time to accelerate to 82.1 m/s, then the x displacement during the time elapsed.
1 answer:
Draw a velocity-time diagram as shown below.
Because a velocity of 26.82 m/s is attained in 4.00 s from rest, the average acceleration is
a = 26.82/4 = 6.705 m/s²
The time required to reach maximum velocity of 82.1 m/s is
t₁ = (82.1 m/s)/(6.705 m/s²) = 12.2446 s
The distance traveled during the acceleration phase is
s₁ = (1/2)at₁²
= (1/2)*(6.705 m/s²)*(12.2446 s)²
= 502.64 m
Answer:
The time required to reach maximum speed is 12.245 s
The distance traveled during the acceleration phase is 502.6 m
Because a velocity of 26.82 m/s is attained in 4.00 s from rest, the average acceleration is
a = 26.82/4 = 6.705 m/s²
The time required to reach maximum velocity of 82.1 m/s is
t₁ = (82.1 m/s)/(6.705 m/s²) = 12.2446 s
The distance traveled during the acceleration phase is
s₁ = (1/2)at₁²
= (1/2)*(6.705 m/s²)*(12.2446 s)²
= 502.64 m
Answer:
The time required to reach maximum speed is 12.245 s
The distance traveled during the acceleration phase is 502.6 m
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