If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1s and 2s is :

Question

If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1s and 2s is : (A) 3 A + 7B (B) (3/2) A + (7/3) B (C) (A/2) + (B/3) (D) (3/2) A + 4 B

Tardigrade Admin · Accepted Answer

V = At + Bt2 ⇒ (dx/dt) =At + Bt2 ⇒ ∫x0 dx = ∫21 (At + Bt2)dt ⇒ x = (A/2)(22 - 12) + (B/3)(23 - 13) = (3A/2) + (7B/3)

Q. If the velocity of a particle is $v = A t + B t^{2}$ , where A and B are constants, then the distance travelled by it between 1s and 2s is :

$3 A + 7 B$

$\frac{3}{2} A + \frac{7}{3} B$

$\frac{A}{2} + \frac{B}{3}$

$\frac{3}{2} A + 4 B$

$V = A t + B t^{2} \Rightarrow \frac{d x}{d t} = A t + B t^{2}$
$\Rightarrow \int_{0}^{x} d x = \int_{1}^{2} (A t + B t^{2}) d t$
$\Rightarrow x = \frac{A}{2} (2^{2} - 1^{2}) + \frac{B}{3} (2^{3} - 1^{3}) = \frac{3 A}{2} + \frac{7 B}{3}$

Q. If the velocity of a particle is v=At+Bt2, where A and B are constants, then the distance travelled by it between 1s and 2s is :

3A+7B

23​A+37​B

2A​+3B​

23​A+4B