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3 years ago

A- what constant acceleration, in SI units, must a car have to gofrom zero to 60 mph in 10 s?

Physics

2 answers:

lorasvet [3.4K]3 years ago

4 0

Answer:

a) acceleration a = 2.68 m/s^2

b) fraction of g = 0.273 or 27.3%

c) distance travelled d = 134m (in SI unit)

d = 439.5ft (in feet)

Explanation:

a) converting 60mph (miles per hour) to m/s

∆v = 60 miles/hour × 1609.344meters/mile × 1/3600 seconds/hour

∆v = 26.82m/s

t = 10s

acceleration = change in velocity/time = ∆v/t

Acceleration a = 26.82/10 = 2.68m/s^2

b) acceleration due to gravity is g = 9.8m/s^2

Fraction of g = a/g = 2.68/9.8 = 0.273

Fraction = 0.273 or 27.3%

c) distance travelled can be calculated using the equation below.

d = ut + 0.5at^2 ....1

Initial speed u = 0

time t = 10s

Acceleration a = 2.68m/s^2

d = distance covered.

From equation 1, putting u = 0

d = 0.5at^2

d = 0.5(2.68 × 10^2)

d = 134m

Converting meters to feet

d = 134m × 3.28ft/m

d = 439.5ft

umka2103 [35]3 years ago

3 0

Answer:

(a) 0.017m/s^2

(b) 17/100,000

Explanation:

(a) speed = 60mph = 60m/1h × 1h/3600s = 0.017m/s, time = 10s

Acceleration (a) = speed ÷ time = 0.017m/s ÷ 10s = 0.0017m/s^2

(b) g = 9.8m/s^2, a = 0.0017m/s^2

a/g = 0.0017/9.8 = 0.00017 = 17/100,000

Distance in foot = 0.17 × 3.2808ft = 0.558ft

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4 years ago

Read 2 more answers

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W = ΔE

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We choose our reference point as a horizontal line at the block's rest point. At the rest, block doesn't have kinetic energy and since it is on the reference point(as we decided) it also has no potential energy.

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