v(t)=A(1−e−t/to)
A(1−e−(2.6/7))=−28
A(0.31)=−28
A=−28/0.31=−90.322m/s
As t tends to 0, (1-e-t/7) tends to 0.So,maximum velocity is given by:
vmax=0m/s ( Can be obtained from graph attached below,t cannot be negative so max value when t is 0)
a(t)=A(e−t/to)/to
a(t)=−90.322(e−t/to)/to
From the graph of acceleration vs time,we can get:
amax=0m/s2as graph converges to 0 as t tends to infinity
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