a. ) In the Daytona 500 auto race , a Ford thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.5m/s . The driver of the Thunderbird realizes he must make a pit stop and he smoothly slows to a stop over a distance of 250m. He spends 5.00s in the pit and then accelerates out , reaching his previous speed of 71.5m/s after a distance of 350m. At this point how far was the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed?
b.) A basketball player who is 2.00m tall is standing on the floor 10.0m . if he shoots the ball at a 40degrees angle with the horizontal at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05m
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- in the daytona 500 auto race a ford thunderbird and a mercedes benz are moving side by side down a straightaway at 71 5 m/s the driver of the thunderbird realizes he must make a pit stop and he smoothly slows to a stop over a distance of 250 m he spends 5
- ?x = v?t 0 5at²
- in the daytona 500 auto race a ford thunderbird and a mercedes benz are moving side by side down a straightaway at 71 5 m/s th
- thunder bird and mercedes physics question
March 21st, 2011 at 2:09 am
a) There may be a more efficient way to do this, but here’s how I would do it. First, find the time it takes the T-bird to slow from 71.5m/s to rest over the 250m. To find the time, you need the rate of acceleration. It is:
v² = v₀² + 2aΔx
Solved for a:
a = (v² - v₀²) / 2Δx
= [0 - (71.5m/s)²] / (2 x 250m)
= -10.2m/s²
The time is found from:
v = v₀ + at
t = (v - v₀) / a
= (0 - 71.5m/s) / -10.2m/s²
= 7.01s
Then, use the same method to find the time the T-bird’s spends accelerating from rest to 71.5m/s over 350m, finding acceleration first:
a = (v² - v₀²) / 2Δx
= [(71.5m/s)² - 0] / (2 x 350m)
= 7.30m/s²
The time here is:
t = (v - v₀) / a
= (71.5m/s - 0) / 7.30m/s²
= 9.79s
Adding all these times up (including the 5.00s pit stop) you get 21.8s. Over that same time the MB moves through a total displacement of:
Δx = v₀t + 0.5at²
= (71.5m/s)(21.8s) + 0——->constant speed (a = 0)
= 1559m
Over that same time the T-bird only moves through a distance of 250m + 350m = 600m, and :
1559m - 600m =959m
b) I guess the player is 10m horizontally from the basket, the problem is not clear. If so, then write an expression for the time for the ball to reach the hoop from the horizontal components of the balls flight:
Δx = (v₀ + v)t / 2
Solved for t:
t = 2Δx / (v₀ + v)
= (2 x 10.0m) / (v₀cos40.0° + v₀cos40.0°)
= 13.1m / v₀
Plugging this into the an aquation for the displacement vertically:
Δy = v₀t + 0.5at²
(3.05m - 2.00m) = v₀sin40.0°( 13.1m / v₀) + 0.5(-9.80m/s²)(13.0m / v₀)²
1.05m = 8.42m - (4.90m/s²)(169m³ / v₀²)
7.42m = 828.1m²/s² / v₀²
v₀ = √[828.1m³/s² / 7.42m]
= 10.6m/s
Hope this helps.