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A box slides down an inclined plane with a uniform acceleration and attains a velocity of 27 m/s in 3 seconds from rest. Find the final velocity and distance moves in 6 seconds (initially at rest)

Respuesta :

The box attains a velocity of 27 m/s in 3 seconds from rest. That means the velocity become 27m/s from 0m/s. Then, the acceleration would be:

acceleration= final velocity - initial velocity / time
acceleration= (27m/s- 0m/s) / 3second= 9m/s2

The final velocity if the time is 6 second would be:
acceleration= final velocity - initial velocity / time
9m/s2= (final velocity - 0m/s) / 6s
final velocity= 9m/s2 * 6s= 54m/s

The distance would be:
distance= (final velocity + initial velocity) * time / 2 
distance= (54m/s+ 0m/s) * 6s/ 2 = 162m
Histrionicus

In the given case, the final velocity after 6s is 54 m/s, and the distance covered by the object in 6s = 162m

Final velocity

To find out the final velocity and distance in 6 seconds first we need to find out the acceleration of the object

Given

initial velocity, u = 0 m/s

final velocity, v = 27 m/s

Time, t = 3 sec

Solution

Using the equation of motion

[tex]v=u+at[/tex]

Putting the values:-

[tex]27= 0+a \times 3\\a=\frac{27}{3}\\= 9\ m/s^2[/tex]

So the acceleration of the object is [tex]9\ m/s^2[/tex]

Now for

t = 6 sec

Using the equation of motion

[tex]v=u+at[/tex]

And using the value of acceleration a = [tex]9\ m/s^2[/tex]

[tex]v=0+9\times6\\= 54\ m/s[/tex]

Now to calculate the distance covered in 6 sec, using the equation of motion

[tex]v^2 -u^2= 2\times a \times s[/tex]

Putting the values

[tex]54^2 -0^2 =2\times 9\times s\\2916=18\times s\\s=\frac{2916}{18} \\=162\ m[/tex]

So, the final velocity after 6s is 54 m/s, and the distance covered by the object in 6s = 162m

Learn more about final velocity:

https://brainly.com/question/13066230

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