Part A At time t = 1 an object moving along a straight line with a constant acceleration of 3m/s2 is located at s(1)=5 and has a velocity of v(1)=10 . What is the equation for the velocity as a function of time? (1 point) Responses v(t)=7+1.5t v ( t ) = 7 + 1.5 t v(t)=7+3t v ( t ) = 7 + 3 t v(t)=10−1.5t v ( t ) = 10 − 1.5 t v(t)=10+1.5t v ( t ) = 10 + 1.5 t v(t)=10+3t v ( t ) = 10 + 3 t Question 2 Part B What is the equation for the location as a function of time? (1 point) Responses s(t)=−2.75+7t+0.75t2 s ( t ) = − 2.75 + 7 t + 0.75 t 2 s(t)=−3.5+7t+1.5t2 s ( t ) = − 3.5 + 7 t + 1.5 t 2 s(t)=5+10t−0.75t2 s ( t ) = 5 + 10 t − 0.75 t 2 s(t)=5+10t+0.75t2 s ( t ) = 5 + 10 t + 0.75 t 2 s(t)=5+10t+1.5t2 s ( t ) = 5 + 10 t + 1.5 t 2 Skip to navigation
Part A: The equation for velocity as a function of time can be found using the formula:
v(t) = v(1) + a(t - 1)
where v(t) is the velocity at time t, v(1) is the initial velocity at time t=1, a is the constant acceleration, and t is the time.
Given that v(1)=10 and a=3, the equation for velocity as a function of time is:
v(t) = 10 + 3(t - 1)
Therefore, the correct response is v(t) = 10 + 3t.
Part B: The equation for location as a function of time can be found using the formula:
s(t) = s(1) + v(1)(t - 1) + 0.5a(t - 1)^2
where s(t) is the location at time t, s(1) is the initial location at time t=1, v(1) is the initial velocity at time t=1, a is the constant acceleration, and t is the time.
Given that s(1) = 5, v(1) = 10, and a = 3, the equation for location as a function of time is:
s(t) = 5 + 10(t - 1) + 0.5(3)(t - 1)^2
Simplifying and expanding:
s(t) = 5 + 10t - 10 + 1.5(t^2 - 2t + 1)
s(t) = 5 + 10t - 10 + 1.5t^2 - 3t + 1.5
s(t) = 1.5t^2 + 7t - 3.5
Therefore, the correct response is s(t) = 1.5t^2 + 7t - 3.5.