a man weighing 80 kg is standing on a trolly weighing320 kg the trolly is resting on a horizontal rails if the man start walking on the trolly along the rails at the speed 1 m/s then after 4s his displacement relative to ground will be
To find the displacement of the man relative to the ground after 4 seconds, we need to consider the relationship between the man, the trolley, and the ground.
Initially, both the man and the trolley are at rest, so their initial velocities are zero. When the man starts walking on the trolley, he imparts a force on it, causing it to move forward. We assume that there is no friction between the trolley and the rails, so the trolley will move with the same velocity as the man.
The force exerted by the man on the trolley can be determined using Newton's second law, which states that force equals mass multiplied by acceleration (F = ma). Since the man and the trolley move together, their combined mass is 80 kg + 320 kg = 400 kg.
Next, we need to calculate the acceleration of the man and the trolley. The acceleration can be found using the formula a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time.
The man's final velocity after 4 seconds is given as 1 m/s, and since there is no initial velocity, the change in velocity is 1 m/s - 0 = 1 m/s. The change in time is given as 4 seconds.
Substituting these values into the formula, we have a = 1 m/s / 4 s = 0.25 m/s^2.
Now that we have the acceleration, we can calculate the displacement of the man and the trolley relative to the ground using the kinematic equation s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration.
Since both the man and the trolley have zero initial velocity, the equation simplifies to s = (1/2)at^2.
Substituting the values, we have s = (1/2)(0.25 m/s^2)(4 s)^2.
Evaluating this expression, we get s = (1/2)(0.25 m/s^2)(16 s^2) = 2 m.
Therefore, the displacement of the man relative to the ground after 4 seconds will be 2 meters.