Question

v = 1 + (t^3)/10 – (t^2)/20
Find the distance travelled from t = 0 and t = 4.

Answer 1

Answer:

130! @+@+2-_+2(384) -64

34

Answer 2

Step-by-step explanation:

Given:

$v = 1 + \frac{ {t}^{3} }{10} - \frac{ {t}^{2} }{20}$

To find: Find the distance travelled from t=0 and t=4.

Solution:

Concept to be implemented:

Rate of change of distance with respect to time is velocity(v)

or

$\bf \frac{dx}{dt} = v$

here, x is distance and y is time.

Write formula in terms of distance.

As, on taking integration; one can find Distance

$x = \int \: v \: dt$

here, distance travelled from t = 0 and t = 4,

So

$x = \int_0^4 \: v \: dt$

$x = \int_0^4 \:\left( 1 + \frac{ {t}^{3} }{10} - \frac{ {t}^{2} }{20}\right)dt$

$x = \:\left( t + \frac{ {t}^{4} }{40} - \frac{ {t}^{3} }{60}\right)\Bigg]_0^4$

Place limits

$x =4 + \frac{ {4}^{4} }{40} - \frac{ {4}^{3} }{60}$

or

$x =4 + \frac{ {256} }{40} - \frac{ 64 }{60}$

or

$x = 4 + 6.4 - 1.067$

or

$x =9 .33$

Final answer:

distance travelled from t = 0 and t = 4 is 9.33 units

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