A car traveling at 28 m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is its acceleration?

2 answers:

6 0

The car's acceleration can be modeled by the following formula:

$a = \frac{v_f - v_i}{t}$

vf represents the final velocity of the car. vi represents the initial velocity of the car. t represents the total time it took for the car to get from its initial to its final velocity.

We already have our values for our velocities and time. Plug them into the equation:

$\text{Initial velocity: 28 m/s}^{2}$
$\text{Final velocity: 0 m/s}^2$
$\text{Time: 7 seconds}$

$\frac{0-28}{7} = -\frac{28}{7} = \boxed{-4}$

The acceleration of the car is -4 m/s, or -4 meters per second.

Alika [10]3 years ago

5 0

The function that models the scenario is
v = init_v + acceleration * time

Plug the values into the equation,
0 = 28 + a * 7 => -28 = a * 7 => a = - 28/7 m/s.

You might be interested in

What is 7.55×8 please show work

iVinArrow [24]

Is this Long multiplication?

7 0

4 years ago

Read 2 more answers

Write an equation of a parabola that passes through (3,-30) and has x-intercepts of -2 and 18. Then find the average rate of cha

Nookie1986 [14]

Answer:

The equation of the parabola is $y = \frac{2}{5}\cdot x^{2}-\frac{32}{5}\cdot x -\frac{72}{5}$ . The average rate of change of the parabola is -4.

Step-by-step explanation:

We must remember that a parabola is represented by a quadratic function, which can be formed by knowing three different points. A quadratic function is standard form is represented by:

$y = a\cdot x^{2}+b\cdot x + c$

Where:

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

$a$ , $b$ , $c$ - Coefficients, dimensionless.

If we know that $(3, -30)$ , $(-2, 0)$ and $(18, 0)$ are part of the parabola, the following linear system of equations is formed:

$9\cdot a +3\cdot b + c = -30$

$4\cdot a -2\cdot b +c = 0$

$324\cdot a +18\cdot b + c = 0$

This system can be solved both by algebraic means (substitution, elimination, equalization, determinant) and by numerical methods. The solution of the linear system is:

$a = \frac{2}{5}$ , $b = -\frac{32}{5}$ , $c = -\frac{72}{5}$ .