A car slows down at -5.00 m/s2 until it comes to a stop after traveling 15.0 m. How much time did it take to stop.

Mathematics

1 answer:

professor190 [17]1 year ago

4 0

Answer:

2.45 seconds to the nearest hundredth.

Step-by-step explanation:

Acceleration a = -5, distance s = 15, time = ?.

Final velocity = v = 0.

We need to use the equations of motion under constant acceleration.

v^2 = u^2 + 2as where u is the initial velocity

0 = u^2 + 2*-5*15

u^2 = 150

so u = √150

Now we use v = u + at:

0 = √150 - 5t

t = √150/5

t = 2.45 seconds.

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Solve the system by graphing. Write the solution as an ordered pair. y = 1/3x + 2 y = –x – 2

Dima020 [189]

ANSWER

The solution is where the two graphs intersect, which is
$(-3,1).$

EXPLANATION

The given system of equations are
$y = \frac{1}{3} x + 2$
and

$y = - x - 2$

We need to graph the two equations.

Let us graph

$y = \frac{1}{3} x + 2$
first.

We need at least two points.

You can choose any appropriate value for x and solve for y. Choosing zero makes our working easier. So let us plot the intercepts.

When
$x = 0$

$\Rightarrow \: y = \frac{1}{3} (0) + 2$

$\Rightarrow y = 0 + 2$

$\Rightarrow y = 2$

So this gives us the ordered pair,

$(0,2)$

When
$y = 0$
we get,

$0= \frac{1}{3} x + 2$

$\Rightarrow \: - 2 = \frac{1}{3} x$

$\Rightarrow \: - 2 \times 3= x$

$\Rightarrow \: -6= x$
This also gives the ordered pair

$(-6,0).$

We plot these two points and draw a straight line through them to obtain the blue graph in the attachment.

For the second line

$y = - x - 2$

We again find the intercepts and plot them.

When
$x = 0$

$y = - 0 - 2$

$\Rightarrow \: y = - 2$

This gives the ordered pair

$(0,-2)$

Also, when
$y = 0$

then we have,

$0 = - x - 2$

$2 = - x$

$x = - 2$

Then we again have the ordered pair,

$(-2,0)$

We plot these two points on the same graph sheet to obtain the red graph above.

The intersection of the two lines is
$(-3,1)$

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