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gulaghasi [49]
3 years ago
12

A driver of a car traveling at 15.0 m/s applies the brakes, causing a uniform acceleration of -2.0 m/s/s. How long does it take

the car to come to a stop? Use the GUESS method. Please help me asap.
Physics
1 answer:
Maksim231197 [3]3 years ago
3 0
For this, we will use the formula:

<span>a = [V₀⁻V₁]÷t

</span>Where:

a - Acceleration
<span>V₀ - Final Velocity
</span>V₁ - Initial Velocity
<span>t - Time

</span>Given:
a = -2.0 m/s²
V₁ = 15.0 m/s
V₀ = 0 m/s ; because the car stopped

Find:
t = ?

Solution:

a = [V₀⁻V₁]÷t

First, we should derive the formula to the element that we are looking for (which is time).

t = [V₀⁻V₁]÷a
t = [0 m/s - 15.0 m/s]÷(-2.0 m/s²) ; substituting the numbers
t = -15.0 m/s ÷ -2.0m/s²
t = 7.5 seconds ; cancelling the removable units thus resulting only to the time unit s(seconds)

It took the car 7.5 seconds to stop
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A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how
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<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity V_{o} has two components, because the brick was thrown at an angle \alpha=61\º:

V_{ox}=V_{o}cos\alpha   (1)

V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}  (2)

V_{oy}=V_{o}sin\alpha   (3)

V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}   (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

X=V_{ox}.t  (5)

Where:

X=10.1m is the distance where the brick landed

t is the time in seconds

If we already know X and V_{ox}, we have to find the time (we will need it for the following equation):

t= \frac{X}{ V_{ox}}  (6)

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<h2>In the y-axis: </h2>

-y=V_{oy}.t+\frac{1}{2}g.t^{2}   (8)

Where:

y is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

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Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

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-y=-10.52m   (10)

Multiplying by -1 each side of the equation:

y=10.52m >>>>This is the height of the building

3 0
3 years ago
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