An object, initially at rest, is subject to an acceleration if 34 m/s^2. how long will it take for that object to travel 3400m
To find the time it takes for an object to travel a certain distance, we can use the following kinematic equation:
\(d = ut + \frac{1}{2}at^2\)
Where:
- \(d\) is the distance traveled (in meters)
- \(u\) is the initial velocity of the object (in meters per second)
- \(a\) is the acceleration of the object (in meters per second squared)
- \(t\) is the time taken (in seconds)
In this case, the object is initially at rest, which means the initial velocity, \(u\), is 0 m/s.
The acceleration, \(a\), is given as 34 m/s^2.
The distance traveled, \(d\), is given as 3400 m.
Substituting these values into the equation, we get:
\(3400 = 0 \cdot t + \frac{1}{2} \cdot 34 \cdot t^2\)
Simplifying the equation, we have:
\(3400 = 17t^2\)
Dividing both sides by 17:
\(t^2 = \frac{3400}{17}\)
\(t^2 = 200\)
Taking the square root of both sides:
\(t = \sqrt{200}\)
Using a calculator, we find:
\(t \approx 14.14\) seconds
Therefore, it will take approximately 14.14 seconds for the object to travel 3400 meters with an acceleration of 34 m/s^2.
To find the time it takes for the object to travel a distance of 3400m with an acceleration of 34 m/s^2, you can use the following equation of motion:
s = ut + (1/2)at^2
where:
s = displacement (3400m)
u = initial velocity (0 m/s since the object is at rest)
a = acceleration (34 m/s^2)
t = time taken
Substituting the known values into the equation:
3400 = 0*t + (1/2)*34*t^2
Rearranging the equation:
3400 = 17t^2
Dividing both sides by 17:
t^2 = 200
Taking the square root of both sides:
t = √200
Calculating the value:
t ≈ 14.1421
Therefore, the object will take approximately 14.1421 seconds to travel a distance of 3400m with an acceleration of 34 m/s^2.