Whereas the gradient of the distance versus time graph gave speed, the gradient of the displacement (a vector) versus time graph gives velocity (a vector). The graph of the velocity v versus time t of a moving object is shown in figure 2 above.Note that this is displacement (a vector quantity) versus time. According to the graph above, uniform acceleration in the intervals (3, 4) and (6, 7). Uniform acceleration happens when the velocity increases linearly with time. Over which time interval(s) was the object accelerating uniformly? The graph of the velocity v versus time t of a moving object is shown in figure 2 above. Note: the distance was found to be 15 m in question 6 above.Īverage speed = distance / time = 15 / 9 = 1.7 m/s (rounded to 2 sf) What is the average speed over the interval t = 0 to t = 9 seconds? Note: the displacement was found to be zero in question 5 above.Īverage velocity = displacement / time = 0 / 9 = 0 What is the average velocity over the interval t = 0 to t = 9 seconds? Total distance from (t= 0 to t = 9) = 7.5 + 7.5 = 15 m The distance is given by the absolute value of the displacement which is given by the area.ĭistance from (t = 6 to t = 9) = | (1/2)(3)(5) | = 7.5 The distance is given by the absolute value of the displacement which is given by the area.ĭistance from (t=0 to t=4) = | -(1/2)(4 + 2)(2.5) | = 7.5īetween t = 6 and t = 9, the object is moving in the positive direction (velocity positive). What is the total distance covered by the object from t = 0 to t = 9 seconds?īetween t = 0 and t = 4, the object is moving in the negative direction (velocity negative).
Total displacement from t= 0 to t = 9 = Total area = -7.5 + 7.5 = 0 The displacement is given by the area between the t-axis and the graph of the velocity.Īrea of Trapezoid on the left below t-axis = -(1/2)(4 + 2)(2.5) = -7.5Īrea of Triangle on the right above t-axis = (1/2)(3)(5) = 7.5 What is the total displacement from t = 0 to t = 9 seconds? Question 5 to 10 refer to the displacement vs time below. In both intervals 0 to 1 and 9 to 11, x is increasing, the velocity is positive and therefore the object is moving in the positive direction. Over which time interval(s) was the object moving in the positive direction? The graph of the position x versus time t of a moving object is shown in figure 1 above. Over which time interval(s) was the object moving in the negative direction?įrom t = 4 s to t = 9 s, x is decreasing and therefore the velocity is negative which means the object is moving in the negative direction.
At what time was the object furthest from the origin (x = 0)?Īt t = 9 s, x = -15 m and this is the furthest point from x = 0. The position of moving object does not change between t = 1 s and t = 4 sĪnd therefore the velocity is equal to zero in this interval of time. On which time interval(s) is the velocity of the moving object equal to zero?
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