Question

A particle’s velocity is given (in units of meters per second) by the function
v = 1 + (t^3)/10 – (t^2)/20
Find the distance travelled from t = 0 and t = 2.

Answer 1

Step-by-step explanation:

S (t)= ∫ v dt

= t + t^4/20 - t^3/60 + k

S(2) = 2 + 16/20 - 8/60 + k

S(0) = k

Distance travelled = S(2) - S(0)

= 2 + 40/60 = 8/3 metres.

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=2.

Solution:

Concept to be implemented:

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

or

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

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 = 2,

So,

$x = \int_0^2 \: v \: dt$

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

$x = \:\left( t + \frac{ {t}^{4} }{40} - \frac{ {t}^{3} }{60}\right)\Bigg]_0^2$

Place limits

$x =2 + \frac{ {2}^{4} }{40} - \frac{ {2}^{3} }{60}$

or

$x =2 + \frac{ {16} }{40} - \frac{ 8 }{60}$

or

$x = 2 + 0.4 - 0.133$

or

$\bf x =2.267$

Final answer:

distance travelled from t = 0 and t = 2 is 2.267 units

Hope it will helps you.

To learn more:

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