Motion in a Straight Line - Result Question 9
9. If the velocity of a particle is $v=At+Bt^{2}$, where $A$ and $B$ are constants, then the distance travelled by it between $1 s$ and $2 s$ is :
(a) $\frac{3}{2} A+4 B$
(c) $\frac{3}{2} A+\frac{7}{3} B$
(b) $3 A+7 B$
(d) $\frac{A}{2}+\frac{B}{3}$
Show Answer
Answer:
Correct Answer: 9. (c)
Solution:
- (c) Given : Velocity $v=At+Bt^{2}$
$\Rightarrow \frac{d x}{d t}=At+Bt^{2}$
By integrating we get distance travelled by the particle between $1 s$ and $2 s$,
$\Rightarrow \int_0^{x} d x=\int_1^{2}(A t+B t^{2}) d t$
$x=\frac{A}{2}(2^{2}-1^{2})+\frac{B}{3}(2^{3}-1^{3})=\frac{3 A}{2}+\frac{7 B}{3}$
