contestada

A car starts from rest and accelerates at a constant rate in a straight line. in the first second the car covers a distance of 2.0 meters. how fast will the car be moving at the end of the second second?

Respuesta :

BlueSky06

The solution would be like this for this specific problem:

Given:

Distance = 2.0 meters

Find:

Vx = ?

 

Δx = [tex]v_ {0x} t + \frac {1} {2}^ {2} ;[/tex]

 

a = 2 Δx [tex]t^{2} = 2 ( \frac {2m} {1s^{2}} ) = 4 \frac {m} {s^{2}} [/tex]

 

[tex]v_ {x} = v_ {0x} + at = ( 4 \frac {m} {s^{2}}) (2 s) = 8 \frac {m} {s}[/tex]

 

Therefore, the car will be moving 8 m/s at the end of the second second.

To add, the physical vector quantity that both requires magnitude and direction is called velocity. It is also equivalent to a direction of motion and specification of its speed.

hamzaahmeds

This question can be solved by the use of equations of motion.

The car will be moving with a velocity of 8 m/s.

First, we will calculate the acceleration of the car by using the second equation of motion on the first second:

[tex]s = v_it+\frac{1}{2}at^2[/tex]

where,

s = distance covered = 2 m

[tex]v_i[/tex] = initial velocity = 0 m/s

t = time = 1 s

a = acceleration = ?

Therefore,

[tex]2\ m = (0\ m/s)(1\ s)+\frac{1}{2}a(1\ s)^2\\\\a = 4\ m/s^2\\[/tex]

Now, we will calculate the final speed of the car after two seconds, by applying the first equation of motion to the complete motion of the car:

[tex]v_f = v_i + at\\\\v_f = 0\ m/s + (4\ m/s^2)(2\ s)\\[/tex]

vf = 8 m/s

The attached picture shows the equations of motion.

Learn more about the equations of motion here:

https://brainly.com/question/20594939?referrer=searchResults

Ver imagen hamzaahmeds

Otras preguntas