a. Average velocity: 15/2=7.5m/s, time= distanc/avgvelociyt= 45/7.5 sec
b. 10 m? at the start, or near the end? Assuming the first 10 m
distance= vi*time-1/2 a t^2
so first determine acceleration: a= changevlocyt/time=-15/(45/7.5)= -2.5m/s^2
10=15*time-2.5/2 time ^2 so a quadratic. In standard form...
1.25t^2-15t +10=0
t= (15+-sqrt(225-50)/2.5=6 +-sqrt175 /2.5= 11.6, 0.709, and reject the 11.6 because it is greater than the time to stop. tme= .709 seconds
a box of lollipops accidentally falls from the back of a truck and hits the ground with a speed of 15 m/s. it slides along the ground for a distance of 45 m before coming to rest. determine
a) the length of time the box slides from stopping
b)the time it takes to slide 10.0 meters
2 answers
oops, i meant the last 10 meters. do i just have to subtract instead of adding the 1/2 a t^2
1 answer