The velocity function (in meters per second) is given for a particle moving alon
ID: 2878360 • Letter: T
Question
The velocity function (in meters per second) is given for a particle moving along a line. v(t) = t^2 - 2t - 24, 1 lessthanorequalto t lessthanorequalto 7 (a) Find the displacement. m (b) Find the distance traveled by the particle during the given time interval. m The acceleration function (in m/s^2) and the initial velocity v(0) are given for a particle moving along a line. a(t) = 2t + 4, v(0) = -5, 0 lessthanorequalto t lessthanorequalto 3 (a) Find the velocity at time t. v(t) = t^2 + 4t - 5 m/s (b) Find the distance travel during the given time interval. 72 mExplanation / Answer
13) given v(t)=t2-2t -24, 1<=t<=7
a) displacement =[1 to 7]v(t) dt
displacement =[1 to 7](t2-2t -24) dt
displacement =[1 to 7]((1/3)t3-t2 -24t)
displacement =((1/3)73-72 -24*7)-((1/3)13-12 -24*1)
displacement =-78 m (negative side indicates left side of initial position)
b)
distance =[1 to 7]|v(t)| dt
distance =[1 to 7]|(t2-2t -24)| dt
v(t)>0
=>t2-2t -24>0
=>(t-6)(t+4)>0
=>t-6>0
=> t>6
v(t)<0
=>t2-2t -24<0
=>(t-6)(t+4)<0
=>t-6<0
=> t<6
when v(t)>0 then |v(t)|=v(t)
when v(t)<0 then |v(t)|=-v(t)
distance =[1 to 6]|(t2-2t -24)| dt+[6 to 7]|(t2-2t -24)| dt
distance =[1 to 6]-(t2-2t -24) dt+[6 to 7](t2-2t -24) dt
distance =[1 to 6]-((1/3)t3-t2 -24t) +[6 to 7]((1/3)t3-t2 -24t)
distance=-((1/3)63-62 -24*6)+((1/3)13-12 -24*1) +((1/3)73-72 -24*7) -((1/3)63-62 -24*6)
distance =266/3 88.667 m
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