Question

An auto travelling along a straight road increases its speed from $30.0m{s}^{-1}$ to $50.0m{s}^{-1}$ in a distance of $180m$. If the acceleration is constant, how much time elapses while the auto moves this distance ?

• A. $6.0s$
• B. $4.5s$
• C. $3.6s$
• D. $7.0s$

Answer 1

Answer: (B)

Let a be constant acceleration of auto.
Here, $u=30m{s}^{-1}$, $v=50m{s}^{-1}$, $S=180m$
As $v^2-u^2=2aS$
${\left(50\right)}^2-{\left(30\right)}^2=2\times a\times 180$
$\therefore a=\frac{1600}{2\times 180}=\frac{40}{9}m{s}^{-2}$
As $S=ut+\frac{1}{2}a{t}^2$
$\Rightarrow 180=30\times t+\frac{1}{2}\times \frac{40}{9}\times t^2$
$\Rightarrow 2t^2+27t-162=0$
Solving this quadratic equation by quadratic formula, we get
$=4.5s$, $-18s$, ($t$ can’t be negative)
$\therefore .t=4.5s$