First, you will focus on the relationship between velocity and position. Recall
ID: 2077647 • Letter: F
Question
First, you will focus on the relationship between velocity and position. Recall that velocity is the rate og change of position (v_x = dx/dt). This means that the velocity is equal to the slope of the position vs. Time graph. Move the person to the position z = -6 m or enter -6.00 in the position box. If you dragged the person to position. click the pause button and then the clear button. Next, drag the person to the sight to roughly z = 6 m and his direction, returning him to the original position at x = -6m. Move the person quickly. about a few seconds for the sound hip. Your plots should look something the those shown below What is the position of the person when t = 1s? Express your answer numerically in meters to one significant figure Notice that since the position is given by z = -4t^2. when the time t - 1x, the position is z - 4(1)^3 m = 4m What is the velocity of the person where t = 1s? Express your answer numerically in meters per second squared to two significant figure Notice that since the position is given by z =(t) -4t^3. the velocity, which is the first derivative of position will respect to time, is given by v_i = dx/dt - 12t^2. So sweet t =1 n, v -m/s when the time t - 1x, the position is z - 4(1)^3 m = 4m What is the velocity of the person where t = 1s? Express your answer numerically in meters per second squared to two significant figureExplanation / Answer
The position is given by:
x = 4t3
and the velocity is the rate of change of distance with time.
v = dx/dt
differentiate x with respect to time to get v
=> v = 3(4)t3-1 = 12t2
at t = 1s, v = 12(1)2 = 12 m/s
to get acceleration, differentiate velocity with respect to time since a = dv/dt
so, a = 2(12)t2-1 = 24t
therefore, the acceleration at t = 1s will be: a = 24(1) = 24 m/s2.
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