step1:
v=0+2a=1, so a = 1/2 km/min^2
s=0+1/4*2^2=1
step2:
s=1*10
step3:
a = -1/3 km/min^2
s = 1*3 - 1/6 * 3^2 = 3/2
PQ = 1 + 10 + 3/2 = 25/2
avg speed = (25/2)/(2+10+3) = 5/6 km/min
there are two different accelerations, shown above. I assume you can convert to m/s^2
A train starts from rest from station P and accelerates uniformly for 2 minutes reaching a speed of 60 km/hr. It maintain this speed 10 minutes and then s uniformly for 3 minutes to comes to rest at station Q find;
a) The distance in PQ in km
b) The average speed of the train
c) The acceleration in m/s2.
1 answer
1 answer