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3 years ago

9

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

r (mph). At full power, the car can accelerate from zero to 29.0 mph in time 1.10 s .A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0 mph in time 1.10 s , how long would it take to go from zero to 58.0 mph ?

1 answer:

bekas [8.4K]3 years ago

6 0

Answer:

2.2 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Converting mph to m/s

$29\ mph=29\times 0.44704=12.96\ m/s$

$58\ mph=58\times 0.44704=25.93\ m/s$

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{12.96-0}{1.1}\\\Rightarrow a=11.78\ m/s^2$

Considering this acceleration to be constant

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{25.93-0}{11.78}\\\Rightarrow t=2.20\ s$

Time it would take to go from zero to 58.0 mph is 2.2 seconds

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3 years ago

Consider the following cyclic process carried out in two steps on a gas. Step 1: 44 J of heat is added to the gas, and 20. J of

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3 years ago

The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,

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Given that

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we know that

$a=\dfrac{dv}{dt}$

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We know that

$v=\dfrac{dx}{dt}$

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Position at t= 2 s:

$\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{2}$

x=2+0.25 ln2

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Position at t= 5 s:

$\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt$

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3 years ago

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mihalych1998 [28]

Answer:

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there, the correct answer will be c= 11.9 yr.

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