A train traveling with a velocity of +35.0 m/s slows as it enters a twin with an acceleration of -1.50 m/s^2 for 10.0s. What is the velocity at the end of 10.0s?
I’m stuck I really need help
in general for constant acceleration a :
v = Vi + a t where Vi is Velocity initial
x = Xi + Vi t + (1/2) a t^2 where Xi is initial displacement x
so
v = 35 - 1.50 * 10
v = 35 - 15
v = 20 m/s
It takes 40 J to push a large box 4 m across a floor. Assuming that the push is in the same direction as the displacement, what is the size of the force on the box?
The work done on an object is given by the product of force and displacement in the direction of the force:
W = F*d*cos(theta)
where W is the work done, F is the force, d is the displacement, and theta is the angle between the force and the displacement.
In this case, the work done is 40 J and the displacement is 4 m, so we can rearrange the equation to solve for the force:
F = W/(d*cos(theta))
Assuming the push is in the same direction as the displacement, theta = 0 and cos(theta) = 1, so:
F = 40 J/(4 m * 1)
F = 10 N
Therefore, the size of the force on the box is 10 N.
Sure! I can help you with that. To find the velocity at the end of 10.0 seconds, we can use the kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity (u) is 35.0 m/s, the acceleration (a) is -1.50 m/s^2, and the time (t) is 10.0 seconds.
Substituting these values into the equation, we get:
v = 35.0 m/s + (-1.50 m/s^2) × 10.0 s
First, let's calculate the product of acceleration and time:
-1.50 m/s^2 × 10.0 s = -15.0 m/s
Now, we can plug this value back into the equation:
v = 35.0 m/s + (-15.0 m/s)
To simplify, we can add the velocities:
v = 35.0 m/s - 15.0 m/s
The final velocity (v) is:
v = 20.0 m/s
Therefore, the velocity of the train at the end of 10.0 seconds is 20.0 m/s.