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2 years ago

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A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t)=

bt2−ct3, where b = 3.00 m/s^2 and c = 0.100 m/s^3.
Required:
a. Calculate the average velocity of the car for the time interval t = 0 to t = 8.0 s.
b. Calculate the instantaneous velocity of the car at the following times:

1. t = 0
2. t = 4.0 s
3. t = 8.0 s

c. How long after starting from rest is the car again at rest?

1 answer:

True [87]2 years ago

6 0

Answer:

(a) average velocity = 17.6 m/s

(b) when t = 0, v = 0

when t = 4, v = 19.2 m/s

when t = 8, v = 28.8 m/s

(c) after starting from rest, the car will be at rest again in 20 s

Explanation:

Given;

x(t)=bt²−ct³, substitute the given values and the equation will become;

x(t)=3t²−0.1t³

(a)average velocity = total distance / total time

total distance, x(t) = 3t²−0.1t³

x(8) = 3t²−0.1t³

X(8) = 3(8)² - 0.1(8)³

X(8) = 140.8 m

total time = 8 s

average velocity = 140.8 / 8

average velocity = 17.6 m/s

(b) instantaneous velocity = dx / dt

dx / dt = 6t - 0.3t²

when t = 0

v = 0

when t = 4 s

v = 6(4) - 0.3(4²) = 19.2 m/s

when t = 8 s

v = 6(8) - 0.3(8²) = 28.8 m/s

(c) the velocity is zero at dx / dt = 0

6t - 0.3t² = 0

t(6 - 0.3t) = 0

t = 0 or 6 - 0.3t = 0

t = 0 or 0.3t = 6

t = 0 or t = 6 / 0.3

t= 0 or t = 20 s

After starting from rest, the car will be at rest again in 20 s

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2 years ago

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When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

$\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}$ .

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was $(63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}$ . Hence, the weight of water that this object displaced would be $47.0 \; {\rm N}$ .

The mass of water displaced would be:

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The volume of that much water (which this object had displaced) would be:

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The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

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Hence, the density of the oil in this question would be:

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(Rounded to $\text{$3$ sig. fig.}$ )

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