Question

Suppose the velocity of a particle is given by the equation, V=m+nt2, where m=10cms and n=2cm/s3.

A, find the change in velocity of the particle in the initial time interval b/n t₁=2sec and t₂=5sec
B, find the average acceleration in this time interval
C, find the instantaneous acceleration at time t₁=2sec​

Answer 1

Hello!

Begin by plugging in the values for m and n. We get the equation for the velocity of the particle to be:

`v(t) = 10 + 2t^2`

A.

To find the change in velocity over the interval (2s ≤ t ≤ 5s), we can simply find the difference in the velocities at these times.

`\Delta v= v_f - v_i`

For this situation:

`\Delta v = v(5) - v(2)`

Substitute these times for 't' into the equation and solve.

`v(5) = 10 + 2(5^2) = 60 (cm)/(s)\\\\v(2) = 10 + 2(2^2) = 18 (cm)/(s)\\\\\Delta v = 60 - 18 = \boxed{42 (cm)/(s)}`

B.

To find the average acceleration, we must take the SLOPE of the velocity function over this interval using the slope formula:

`a_(avg) = (v_f - v_i)/(\Delta t)`

Plug in the values for the particle's velocity at t = 2 s and 5 s that we solved for above.

`a_(avg) = (60- 18)/(5 - 2)\\\\a_(avg) = (42)/(3) = \boxed{ 14 (cm)/(s^2)}`

C.

The instantaneous acceleration can be found by taking the derivative of the v(t) function using the power rule. Recall:

`(dy)/(dx) x^n = nx^(n-1)`

Using this rule:

`a(t) = v'(t) = 2(2t) = 4t`

Substituting in t = 2 s:

`a(2) = 4(2) = \boxed{8 (cm)/(s^2)}`