106k views
2 votes
Thing 1 pushes Thing 2 from rest up the hall 0.050km with an acceleration of 0.950 m/s. How long did this action of bullying last?​

User Eddy Pronk
by
8.6k points

1 Answer

2 votes

Final answer:

The time it took for 'Thing 1' to push 'Thing 2' with an acceleration of 0.950 m/s over a distance of 0.050 km (50 meters) is approximately 4.58 seconds.

Step-by-step explanation:

The student is asking how long it took for 'Thing 1' to push 'Thing 2' given an acceleration and distance. To find the time, we can use the kinematic equation d = Vi * t + (1/2) * a * t^2, where d is the displacement, Vi is the initial velocity, a is the acceleration, and t is the time.

Given that Thing 2 starts from rest (Vi = 0) and the displacement d is 0.050 km (which is 50 meters), and the acceleration a is 0.950 m/s^2, we can simplify the equation to d = (1/2) * a * t^2.

Plugging in the values we get 50 m = (1/2) * 0.950 m/s^2 * t^2. To solve for t, we first multiply both sides by 2 and then divide by the acceleration, yielding 100 m = 0.950 m/s^2 * t^2. Next, to find t, we take the square root of both sides, which gives us t ≈ 4.58 seconds. This is the duration of the push.

User ZeroNine
by
7.4k points