A constant force F is applied to a body of mass m that initially is headed east at velocity v₀ until its velocity becomes $\Delta$v₀.The total time of travel is 2t.The total distance the body travels in that time is A) $\frac { 1 } { 2 } \frac { F } { m } t ^ { 2 }$ . B) $\frac { F } { m } t ^ { 2 }$ . C) $v _ { 0 } t - \frac { 1 } { 2 } \frac { F } { m } t ^ { 2 }$ . D) $v _ { 0 } t + \frac { 1 } { 2 } \frac { F } { m } t ^ { 2 }$ . E) $2 \left( v _ { 0 } t + \frac { 1 } { 2 } \frac { F } { m } t ^ { 2 } \right)$ .

Question

Quizplus · Accepted Answer

Since the force F is constant, we can use the kinematic equation:

v = u + at

where v is final velocity, u is initial velocity, a is acceleration, and t is time. When the body stops, we have v = 0, u = vS1 , and a = -F/m (because the force is opposite to the initial velocity). Thus,

t = vS1 / (F/m) = mvS1 / F

The distance traveled during the deceleration is given by:

s1 = u*t + 1/2 * a * t^2 = vS1 * mvS1 / F - 1/2 * F/m * (mvS1/F)^2 = (m*vS1^2)/(2F)

The distance traveled during the acceleration from rest to F1 is identical, so the total distance traveled is

s1 + s2 + s3 = 2*(m*vS1^2)/(2F) + (F/m) * (mvS1/F)^2 = 3m*vS1^2/(2F)

Substituting the given values, we obtain:

s1 + s2 + s3 = 3*10/(2*4) = 7.5

Therefore, the best choice is B, 11ea99f6_d74f_8a67_8bb6_ffe2b5653939_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11 .

A Constant Force F Is Applied to a Body of Mass