Question

The Velocity of a particle is given by V= t2-6t+7 Find the distance
covered by 3 seconds.
O A 4 sq units
O B 7 sq.unitd
O C 5 sq.units
D3sq,units​

Answer 1

Answer:

d

Step-by-step explanation:

3śq , units 7+2=9 - 6= 3 Sq units

Answer 2

Answer:

Option D is the correct option.

Step-by-step explanation:

Given that Velocity = V = t² - 6t + 7

We know that Velocity = $\frac{Distance}{Time}$

Since $\frac{d}{dx} (Distance)$ = Velocity

=> $\int (Velocity)$ = Distance

Therefore, Distance = $\int$(t² - 6t + 7)

                                 = $\frac{t^3}{3}$ - $\frac{6t^2}{2}$ + 7t + c

At t = 0 sec, Distance = 0

=> 0 =  $\frac{0}{3}$ - $\frac{6*0}{2}$ + 7*0 + c

=> c = 0

Therefore, distance = $\frac{t^3}{3}$ - $\frac{6t^2}{2}$ + 7t

                                 = $\frac{t^3}{3}$ - 3t² + 7t

We need to find distance at t = 3

=> distance = $\frac{3^3}{3}$ - 3*3² + 7*3

                   = 9 - 27 + 21

                   = 3

Therefore, the distance is 3 units

Therefore, option D is the correct option.