Question #111386
A particle moves along the X axis. Its position varies with time according to the function X(t) = 4t – 2t2, where X is in meters and t is in seconds.

1. Calculate the velocity of the particle V(t)
my answer was (0s,3s)????

2.) Calculate the acceleration of the particle a(t)
My answer was (-2,2s)???

3.) Calculate the velocity at t=2.5 s
dont get that one

4) Calculator the position at time t=3.5
dont get this either

thank you
1
Expert's answer
2020-04-22T10:01:47-0400

1).Speed is the time derivative of the equation of motion

V(t)=dx(t)dt=d(4t2t2)dt=44tm/sV(t)=\frac{dx(t)}{dt}=\frac{d(4t-2t^2)}{dt}=4-4t m/s

i.e

V(t)=44tm/sV(t)=4-4t m/s

2).acceleration is the time derivative of speed

a(t)=dV(t)dt=d(44t)dt=4m/s2a(t)=\frac{dV(t)}{dt}=\frac{d(4-4t)}{dt}=-4 m/s^2

i.e

a(t)=4m/s2a(t)=-4 m/s^2

3.) Speed at t = 2.5 s

V(2.5)=442.5=410=6m/sV(2.5)=4-4 \cdot 2.5=4-10=-6 m/s

4) The position of the particle at time t = 3.5 s

x(3.5)=4t2t2=43.523.52=1424.5=10.5mx(3.5)=4t-2t^2=4 \cdot 3.5-2 \cdot 3.5^2=14-24.5=-10.5 m


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