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ArbitrLikvidat [17]
3 years ago
11

A car starting from rest accelerates in a straight line at a constant rate of 5.5m/s for 6s.If the car after this acceleration s

lows down uniformly at a rate of 2.4m/s, how long does it take to stop​
Physics
1 answer:
aalyn [17]3 years ago
7 0

Answer:

The time it takes to stop is 13.75 seconds

Explanation:

A body moving with constant acceleration, 'a', for a time, 't', has a final velocity, 'v', given by the following kinematic equation;

v = u + a·t

Where;

v = The final velocity of the body

a = The acceleration of the body

t = The time of acceleration (accelerating period) of the body

u = The initial velocity of the body

The given parameters for the acceleration of the car are;

The initial velocity of the car, u = 0 m/s (a car starting from rest)

The constant acceleration of the car, a =  5.5 m/s²

The acceleration duration, t = 6 s

Therefore, we have;

The final velocity of the car after the acceleration, v = 0 m/s + 5.5 m/s² × 6 s = 33 m/s

The final velocity of the car after the acceleration, v = 33 m/s

When the car slows down uniformly, and comes to a stop (final velocity, v₂ = 0 m/s), it has a constant negative acceleration, (deceleration) '-a₂'

The given parameters when the car slows down  are;

The deceleration, -a₂ = 2.4 m/s²

The final velocity, v₂ = 0 m/s

The initial velocity, u₂ = v = 33 m/s

The time it takes to stop = t₂

-a₂ = 2.4 m/s²

∴ a₂ = -2.4 m/s²

From, v = u + a·t, we have;

v₂ = v + a₂·t₂

By plugging in the values of the variables, we have;

0 m/s = 33 m/s + (-2.4 m/s²) × t₂

∴ 2.4 m/s² × t₂ = 33 m/s

t₂ = 33 m/s/(2.4 m/s²) = 13.75 s

The time it takes to stop, t₂ = 13.75 seconds

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If the scaled-up man now stands on one leg, what fraction of the tensile strength is the stress on the femur?.
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<h3>Cross sectional area and stress on the femur</h3>

Question Parameters:

To see why this must be so, recall, that the stress on the femur for a man standing on one leg is 1.4%

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Complete Question

Larger animals have sturdier bones than smaller animals. A mouse's skeleton is only a few percent of its body weight, compared to 16% for an elephant. To see why this must be so, recall, that the stress on the femur for a man standing on one leg is 1.4% of the bone's tensile strength. Suppose we scale this man up by a factor of 10 in all dimensions, keeping the same body proportions. (Assume that a 70 kg person has a femur with a cross-section area (of the cortical bone) of 4.8 x 10-4m², a typical value.)

Part A

Both the inside and outside diameter of the femur, the region of cortical bone, will increase by a factor of 10. What will be the new cross-section area? Express your answer using two significant figures.

Part B

If the scaled-up man now stands on one leg, what fraction of the tensile strength is the stress on the femur?

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