11 views
A point moves in a straight line under the retardation a $v^{2}$, where ' $a$ ' is a positive constant and $v$ is speed. If the initial speed is $u$, the distance covered in ' $t$ ' seconds is:

(A) a ut

(B) $\frac{1}{a} \log (a u t)$

(C) $\frac{1}{a} \log (1+a u t)$

(D) $a \log (a u t)$

The retardation is given by $\frac{d v}{d t}=-a v 2$ integrating between proper limits

$$\begin{array}{l} \Rightarrow \quad-\int_{u}^{v} \frac{d v}{v^{2}}=\int_{0}^{t} a d t \\ \text { or } \quad \frac{1}{v}=a t+\frac{1}{u} \\ \Rightarrow \quad \frac{d t}{d x}=a t+\frac{1}{u} \Rightarrow d x=\frac{u d t}{1+a u t} \end{array}$$

integrating between proper limits

$$\begin{array}{ll} \Rightarrow & \int_{0}^{s} d x=\int_{0}^{1} \frac{u d t}{1+a u t} \\ \Rightarrow & S=\frac{1}{a} \ln (1+a u t) \end{array}$$
by
3.3k Points

## Related Questions

1 Answer 4 Views
4 views
1 Answer 9 Views
9 views
1 Answer 7 Views
7 views
1 Answer 12 Views
12 views
1 Answer 12 Views
12 views
1 Answer 6 Views
6 views