position and movement 2題

Larry

2009-10-22 4:48 am

Q1.A minibus accelerates from rest at 3ms^-2 until it reaches a speed of
15ms^-1.It maintains its speed for 10s and then decelerates to a stop at
1.5ms^-2.Whatt is the total time o travel?

Q2.A skater initially skating along a straight line at 2ms^-1 has an acceleration
of 0.8ms^-2 in the opposite direction. When will he reach a speed of 2ms^-1
again?

回答 (1)

天同

2009-10-22 5:12 am

✔ 最佳答案

Q1: During the acceleration period,
using equation of motion: v = u + a.t
where u = 0 m/s, a = 3 m/s2, v = 15 m/s, t = t1, say
hence, 15 = 3 x t1
t1 = 5 s
During the decceleration period,
use equation of motion: v = u + at
v = 0 m/s, u = 15 m/s, a = -1.5 m/s2, t = t2, say
hence, 0 = 15 + (-1.5).t2
t2 = 15/1.5 s = 10 s

Therefore, the total time of travel = (5 + 10 + 10) s = 25 s

Q2: Use equation of motion: v = u + at
with u = 2 m/s, v = - 2m/s, a = -0.8 m/s2, t = ?
hence, -2 = 2 + (-0.8).t
i.e. t = 4/0.8 s = 5 s

👍 0 👎 0