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.

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

B.

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

$a_{avg} = \frac{v_f - v_i}{\Delta t}$

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

$a_{avg} = \frac{60- 18}{5 - 2}\\\\a_{avg} = \frac{42}{3} = \boxed{ 14 \frac{cm}{s^2}}$

C.

The instantaneous acceleration can be found by taking the derivative of the v(t) function using the power rule. Recall:
$\frac{dy}{dx} x^n = nx^{n-1}$

Using this rule:
$a(t) = v'(t) = 2(2t) = 4t$

Substituting in t = 2 s:
$a(2) = 4(2) = \boxed{8 \frac{cm}{s^2}}$