5. &s=2.00 km | 0 (km) -470 km =670 km Ar\'=-1.50 km x Ckm) xj3.75 k525 km i. Wh

Question

5. &s=2.00 km | 0 (km) -470 km =670 km Ar'=-1.50 km x Ckm) xj3.75 k525 km i. What are the magnitude and sign of displacements for the motions of the subway train shown in parts a and b above? (a) Solve for displacement in part (a). (a) Solve for displacement in part (b). Suppose the train shown above accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval? (mixed units so convert everything to m/s) suppose that at the end of its trip, the train slows to a stop from a speed of 30.0 km/h in ili. Now 8.00 s. What is its average acceleration while stopping?

Explanation / Answer

ii)

v = 30 kmph = 30 x 5/18 m/s = 25/3 = 8.33 m/s ; u =0

a = (v-u)/t = 8.33/20 = 0.4165 m/s^2

same as graph

iii)

a = (0-8.33)/8 = -1.04 m/s^2

same as graph

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