Solving for Distance, Velocity and Acceleration

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The discussion revolves around calculating distance, velocity, and acceleration using the equations x=3e^(0.4t), v=dx/dt, and a=dV/dt. For t=5, the distance is calculated as 22.17m, the velocity as 8.867m/s, and the acceleration as 3.547 m²/s. There is a concern about the dimensional aspects of the equations, which were initially overlooked. Adding units to the time constant clarifies the calculations and ensures dimensional consistency. Overall, the calculations are confirmed to be correct once the units are properly incorporated.
jenny121
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Homework Statement
The distance moved by a mass is related to the time by x=3e^(0.4t) m. Find the following value after 5 sec
Relevant Equations
x=3e^(0.4t)
Distance:
substitute t=5 into x=3e^(0.4t)
22.17m

Velocity:
v=dx/dt
=1.2e^0.4t____(1)
Sub t=5 back into (1)
v= 8.867m/s

Acceleration:
a=dV/dt
=0.48e^0.4t____(2)
sub t=5 back into (2)
a= 3.547 m2/s

I am not sure if i am doing this right on dx/dt and dv/dt
 
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Your calculus looks fine. You might question the dimensional aspects of these equations, which the problem setter has chosen to ignore.
 
PeroK said:
Your calculus looks fine. You might question the dimensional aspects of these equations, which the problem setter has chosen to ignore.
It can be fixed by giving the time constant a unit:
##x=(3m)e^{(0.4s^{-1})t}##.
Those units then flow through to the answer.
 
I see, thanks. after putting the unit it makes more sense now
 
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