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.

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

## Comments

Assignment Expert18.09.19, 16:48Dear visitor,

please use panel for submitting new questions

Abd17.09.19, 21:24What is the particle position at 2s

Assignment Expert13.09.19, 17:22Dear Eman, it is the properties of derivation: (40 - 5*t^2)' = (0 + 2*5*t) = 10t

Eman13.09.19, 16:05Why the ax= -10 and not -20

Please explain

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