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3 years ago

Need 021-028 please.

Physics

1 answer:

larisa86 [58]3 years ago

3 0

(021) The acceleration at 1 second is the slope of the line plotted over the first 2 seconds. This line passes through the points (0 s, 0 m/s) and (2 s, 5 m/s), so its slope and thus the (average) acceleration is

(5 m/s - 0 m/s) / (2 s - 0 s) = 5/2 m/s² = 2.5 m/s²

(022) At time 2 seconds, the velocity curve passes through the point (2 s, 5 m/s), so the velocity is 5 m/s

(023) The distance traveled by this object after 2 seconds is equal to the area under the velocity function over the first 2 seconds. This region is a triangle with base 2 s and height 5 m/s, so the area is

1/2 (2 s) (5 m/s) = 5 m

The object's initial position is 10 m, so its final position after 2 seconds is

10 m + 5 m = 15 m

(024) Similar to (021): compute the slope of the line connecting the point (2 s, 5 m/s) and (6 s, 7 m/s):

(7 m/s - 5 m/s) / (6 s - 2 s) = (2 m/s) / (4 s) = 1/2 m/s² = 0.5 m/s²

(025) We know the distance traveled in the first 2 seconds. Over the next 4 seconds, the object travels an additional distance equal to the area of a trapezoid with "height" 6 s - 2 s = 4 s, and "bases" 5 m/s and 7 m/s. The area of this trapezoid is

1/2 (4 s) (5 m/s + 7 m/s) = 24 m

The net distance traveled after 6 s is then

5 m + 24 m = 29 m

so the object's position after 6 s is

10 m + 29 m = 39 m

(026) Again, compute the slope of a line, this time through the point (6 s, 0 m/s) and (9 s, -1 m/s) :

(-1 m/s - 0 m/s) / (9 s - 6 s) = (-1 m/s) / (3 s) ≈ -0.333 m/s²

(027) If the velocity at 6 s is 0 m/s, and after each second the velocity decreases by 0.333 m/s, then after 2 more seconds the velocity would be

0 m/s + (2 s) (-0.333 m/s²) = -0.666 m/s

(028) The distance traveled between the 6th second and 8th second corresponds to the area of another trapezoid, with "height" 8 s - 6 s = 2s and "bases" 0 m/s and 0.666 m/s - 0 m/s = 0.666 m/s. So the distance traveled in this interval is

1/2 (2 s) (0 m/s + 0.666 m/s) = 0.666 m

However, the velocity is negative for this duration, so the object has turned around and is moving in the opposite direction. This means it covers a net distance after a total of 8 s of

5 m + 24 m - 0.666 m = 28.333 m

and so its position after 8 s is

10 m + 28.333 m = 38.333 m

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