Question

The coordinates of an object is given as a function of time by x=4t^2-3t^3,where x is in meter and t is in second .what is its average acceleration over the time interval from 0 to 2 seconds

Answer 1

Answer:

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Answer 2

Given:

• The displacement versus time function is given as :

$x = 4 {t}^{2} - 3 {t}^{3}$

To find:

• Average acceleration from t = 0 to 2 seconds?

Calculation:

Average acceleration will be calculated as the change in velocity divided by change in time. So, first lets find out the velocity function :

$x = 4 {t}^{2} - 3 {t}^{3}$

$\implies \: v = \dfrac{dx}{dt}$

$\implies \: v = \dfrac{d(4 {t}^{2} - 3 {t}^{3}) }{dt}$

$\implies \: v = 8t - 9 {t}^{2}$

Now, we can say:

• at t = 0 sec , v = 0 m/s.

• at t = 2 sec , v = 8(2) - 9(2²) = 16 - 36 = -20 m/s.

So, average acceleration is :

$avg. \: acc = \dfrac{ - 20 - 0}{2 - 0}$

$\implies avg. \: acc = \dfrac{ - 20 }{2 }$

$\implies avg. \: acc = - 10 \: m {s}^{ - 2}$

So, average acceleration is -10 m/s².