An object is at x = 0 at t = 0 and moves along the x axis according to the veloc

Question

An object is at x = 0 at t = 0 and moves along the x axis according to the velocity-time graph shown below 20 10 t(s) -10 (a) What is the object's acceleration between O and 4.0 s? mys2 (b) What is the object's acceleration between 4.0 s and 9.0 s? m/s2 (c) What is the object's acceleration between 13.0 s and 18.0 s? m/s (d) At what time(s) is the object moving with the lowest speed? (If there is more than one time, enter the times from smallest to largest. If there is only one time, enter 0 in the second box.) (e) At what time is the object farthest from x 0? () What is the final position x of the object at t- 18 ? 9) Through what total distance has the object moved between t-o andt18 s?

Explanation / Answer

(A) slope of v vs t curve gives the acceleration.

from 0 to 4s:

a = slope = 0  

Ans: 0

(B) 4 to 9 s:

a = (18 - (-12))/(9 - 4)

a = 6 m/s^2

Ans: 6

(C) 13 to 18s.

a = (0 - 18) / (18 - 13) = - 3.6 m/s^2

Ans: - 3.6

(D) slowest speed is zero.

t = 6s, 18s

Ans: 6

18

(E) at t = 18 s

Ans: 18

(F) area under the v vs t cruve will give displacement.

delta(x) = xf - xi = (-12 x 4) + (-12 x 2 / 2) + (3 x 18 / 2) + (18 x 4) + (5 x 18 / 2)

x - 0 = 84

x = 84 m  

(g) distance = (12 x 4) + (12 x 2 / 2) + (3 x 18 / 2) + (18 x 4) + (5 x 18 / 2)

= 204 m

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