Question

A uniformly accelerated bus passes over a straight bridge of length 4.2km. With what velocity should the bus enter the bridge with a velocity of $29{m}/{s}\;$ in 4 min?
$\begin{align}
  & 1)2{m}/{s}\; \\
 & 2)4{m}/{s}\; \\
 & 3)6{m}/{s}\; \\
 & 4)8{m}/{s}\; \\
\end{align}$

Answer 1

Hint: Apply the equations of motion to find the solution of this question. That is, first use the first equation of motion connecting the final velocity, initial velocity, acceleration and time. Then apply the distance formula. Then by solving the two equations we will get the initial velocity.

Complete answer:
Given that,
The length of the bridge, $l=4.2km$
$=4.2\times 1000m$
$=42000m$
The final velocity, ${{V}_{f}}=29{m}/{s}\;$
Time, $t=4\min $
$=4\times 60\sec $
$=240\sec $
Let the initial velocity be ${{V}_{i}}$ .
Then by the equations of motion we have,
${{V}_{f}}={{V}_{i}}+at$
Substituting the values we get,
$29={{V}_{i}}+a\times 240$ ………(1)
Again applying the second equation of motion we get,
$l={{V}_{i}}t+\dfrac{1}{2}a{{t}^{2}}$
Substituting the values we get,
$4200={{V}_{i}}\times 240+\dfrac{1}{2}a\times {{(240)}^{2}}$ ………(2)
Multiply equation (1) by we get,
$\dfrac{240}{2}\times 29=\dfrac{240}{2}\times {{V}_{i}}+a\times \dfrac{{{240}^{2}}}{2}$ ……(3)
Subtracting equation (2) by (3) we get,
$720=120{{V}_{i}}$
Rearranging the equation we get,
${{V}_{i}}=\dfrac{720}{120}$
$\therefore {{V}_{i}}=6{m}/{s}\;$

Note: The equations of the motion are very important in solving the problems in kinematics. There are three equations of motion. The variables involved in this equation are distance or displacement, velocity, acceleration and time. These equations are used in the case of uniform or constant acceleration.