Asked by jayven march
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
a) the length of time the box slides from stopping
b)the time it takes to slide 10.0 meters
Answers
Answered by
bobpursley
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
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
Answered by
jayven march
oops, i meant the last 10 meters. do i just have to subtract instead of adding the 1/2 a t^2
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