Question

The position of particle moving along a straight line is related to time as x=4-3t+2t2. The average velocity over a time interval of 1 to 4 second is

7 m/s

10 m/s

15 m/s

8 m/s​

Answer 1

Explanation:

when time = 1 sec

x = 4 - 3(1) + 2(1)²

x = 4 -3 +2

x = 3m/s

when time = 4sec

x = 4 - 3(4) + 2(4)²

x = 4 - 12 + 32

x = 24 m/s

average velocity = (24-3) / (4-1)

= 21/3

= 7 m/s

Answer 2

Answer:

Given:

x = 4 - 3t + 2t²

Time interval = 1 to 4 seconds

To find:

Average velocity in the given interval

$\sf{\underline{Solution:-}}$

☘Velcity at any given time can be found using the differential equation :

$\boxed{v = \dfrac{dx}{dt}}$

✦Putting x = 4 - 3t + 2t² :

➫$v = \dfrac{d}{dt} (4 - 3t + 2t²)$

☘ Using the differentiation of power as :

$\boxed{\dfrac{d}{dx} ax^n = nax^{n-1}}$

✦Now solving the equation:

➫$v = \dfrac{d}{dt} 2t² + \dfrac{d}{dt} -3t + \dfrac{d}{dt} 4$

➫$v = 2 \times 2t^{2‐1} + 1 \times -3t^{1-1} + 0$

➫$v = 4t - 3$

⚕ Velocity at t = 1 :

➫$v = 4(1) - 3$

➫$v = 4 - 3$

➫$v = 1 \ m/s$

⚕ Velocity at t = 4:

➫$v = 4(4) - 3$

➫$v = 16 - 3$

➫$v = 13 \ m/s$

☘ Using average velocity formula :

$\boxed{v_{av} = \dfrac{v_i + v_f}{2}}$

✦Putting the values in this equation :

➫$v_{av} = \dfrac{13 + 1}{2}$

➫$v_{av} = \dfrac{14}{2}$

➫$v_{av} = 7 \ m/s$