A net horizontal force F =A +Bt^3 acts on a 2kg object, where A = 5N and B = 2N/s3. What is the horizontal velocity of this object 4.0 seconds after it starts from rest?
F=ma=5+2t^3
a=2.5+t^3
V=2.5t+(1/4)(t^3)
therefore, at t=4, V=74m/s
Is this correct?
V=2.5t+t^4
To determine the horizontal velocity of the object 4.0 seconds after it starts from rest, we can use the equation V = 2.5t + (1/4)(t^3), where V represents the velocity and t represents the time.
By substituting t = 4 into the equation, we can calculate the velocity:
V = 2.5(4) + (1/4)(4^3)
V = 10 + (1/4)(64)
V = 10 + 16
V = 26 m/s
Therefore, the correct answer is V = 26 m/s, not 74 m/s as you had calculated.