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Answer to Question #21473 in Mechanics | Relativity for faisl
Question #21473
The velocity of a particle moving along the x-axis varies in
time according to the expression vx(t) = (40 – 5 t2) m/s, where t is in
seconds.
Find the average acceleration in the time interval t=0 to t=2.0s.
Determine the acceleration at t = 2.0 s.
time according to the expression vx(t) = (40 – 5 t2) m/s, where t is in
seconds.
Find the average acceleration in the time interval t=0 to t=2.0s.
Determine the acceleration at t = 2.0 s.
Expert's answer
Acceleration is the rate at which the velocity of a body changes with time:
a = dv/dt
a - acceleration
v - velocity
In our case:
ax = dvx(t)/dt
ax = d/dt (40 - 5 t2) = -10t
the average acceleration in the time interval t=0 to t=2.0s:
a_av = [ax(0) + ax(2)]/2 = (0-20)/2 = -10m/s^2
the acceleration at t = 2.0 s
ax(2) = -10*2 = -20 m/s^2
a = dv/dt
a - acceleration
v - velocity
In our case:
ax = dvx(t)/dt
ax = d/dt (40 - 5 t2) = -10t
the average acceleration in the time interval t=0 to t=2.0s:
a_av = [ax(0) + ax(2)]/2 = (0-20)/2 = -10m/s^2
the acceleration at t = 2.0 s
ax(2) = -10*2 = -20 m/s^2
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Comments
Dear visitor, please use panel for submitting new questions
What is the particle position at 2s
Dear Eman, it is the properties of derivation: (40 - 5*t^2)' = (0 + 2*5*t) = 10t
Why the ax= -10 and not -20 Please explain
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