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Anit [1.1K]
2 years ago
11

a sprinter running a 100m dash leaves the starting block and accelerates to a maximum velocity of 11m/s at 6s into the race. the

sprinter maintains this velocity for 2s , and then slows down until crossing the finish line 11s after beginning the race. a) what was the sprinter's average acceleration during the first 6s of the race? b) what was the sprinter's average acceleration from 6 to 8 s into the race?
Physics
1 answer:
mihalych1998 [28]2 years ago
7 0

Answer:

Part(a): The average acceleration is 1.83~m~s^{-2} during the first 6s.

Part(b): The average acceleration from 6s to 8s is zero.

Explanation:

Part(a):

The average acceleration is defined as the rate at which velocity is changing with respect to time. So if 'v_{i}', 'v_{f}' and 'a_{av}' represents the initial velocity, final velocity, and average acceleration of a particle, then mathematically

a_{av} = \dfrac{v_{f} - v_{i}}{t}

where 't' is the time taken by the particle to achieve the velocity 'v_{f}' starting from initial velocity 'v_{i}'

Given in the problem, v_{i} = 0, v_{f} = 11~m~s^{-1}~and~t = 6~s.

So the average acceleration(a_{av}) during the first 6 s will be

a_{av} = \dfrac{11 - 0}{6}~m~s^{-2} = 1.83~m~s^{2}

Part(b):

During the time between 6 s to 8 s, as mentioned in the problem, the sprinter maintains the constant velocity. So the average acceleration during this time interval will be zero.

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Suppose that a Ferrari and a Porsche begin a race with a moving start, and each moves with constant speed. One lap of the track
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Answer:

The speed of the Porsche =  180 km/h

The speed of the Ferrari =  200 km/h

Explanation:

The distance of one lap = 2 km

The distance the Ferrari laps the Porsche = 9 laps

The distance the Ferrari will lap the Porsche with 10 km/h less speed = 18 laps

Let the speed of the Ferrari = S_F<em>, </em>the speed of the Porsche = S_P<em>, </em>and the time that lapse before two cars lapped = t

we have;

S_F × t₁ - S_P × t = 2 km

S_P × t₁ = 9 laps × 2 km/lap = 18 km

S_P × t₁ = 18 km

S_F × t₁ =  2 km + S_P × t = 2 km + 18 km = 20 km

S_F × t₁ =  20 km

∴

Similarly, we are given that the following relation;

(S_F - 10) × t₂ - S_P × t₂ = 2 km...(2)

From which we have;

S_P × t₂ = 18 laps × 2 km/lap = 36 km

S_P × t₂ = 36 km

(S_F - 10) × t₂ = 2 km + 36 km = 38 km

(S_F - 10) × t₂ = 38 km

Therefore, given that S_P × t₂ (36 km) = 2 × S_P × t₁ (18 km), we have;

S_P × t₂ = 2 × S_P × t₁

t₂ = 2×t₁

Equation (2) becomes;

(S_F - 10) × 2×t₁ - S_P × 2×t₁ = 2 km...(2)

From which we have;

(S_F - 10) × 2×t₁  = 38 km

(S_F - 10) × 2×t₁  = 38 km

2 × t₁ × S_F - 2 × t₁ × 10 = 38 km

∵ S_F × t₁ =  20 km, we have;

2 × 20 km - 2 × t₁ × 10 = 38 km

2 × t₁ × 10 = 2 × 20 km - 38 km = 2 km

20 × t₁ = 2 km

t₁ = 2/20 = 0.1 hour = 6 minutes

Therefore, we have;

S_P × t₁ = 18 km

S_P × 0.1 h = 18 km

S_P  = 18 km/0.1 h = 180 km/h

The speed of the Porsche = S_P = 180 km/h

S_F × t₁ =  20 km

S_F × 0.1 h =  20 km

S_F  =  20 km/0.1 h = 200 km/h

The speed of the Ferrari = S_F = 200 km/h

7 0
2 years ago
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