Question

A car moving with constant acceleration covered the distance between two points $60.0m$ apart in $6.00s$ . Its speed as it passed the second point was $15.0m/s$ . What was the speed at the first point?

Answer 1

Hint: Here, we are given the final speed, distance covered, and the time taken to cover the distance. So, using the two equations, $v=u+at$ and ${{v}^{2}}={{u}^{2}}+2as$ , we can find the speed at the first point, i.e., the value of $u$.

Complete step by step solution:
We are given, $v$ = speed of the car at the second point $=15m/s$ ,
$s$ = distance between the two points $=60m$ ,
$t$ = time of motion $=6s$
Now, let, $u$= speed of the car at the first point,
And $a$ = acceleration.
From the equation of motion, we know, $v=u+at$
So, $15=u+6\times a$
Therefore, $a=\dfrac{15-u}{6}$ ……………………(i)
Now, we also know that, ${{v}^{2}}={{u}^{2}}+2as$
Therefore, using the value of $a$ from the equation (i), we can have,
$\Rightarrow 225={{u}^{2}}+\dfrac{(15-u)}{6}\times 60$
$\Rightarrow 225={{u}^{2}}+300-20u$
$\Rightarrow {{u}^{2}}+75-20u=0$
$\Rightarrow {{u}^{2}}-15u-5u+75=0$
Now, solving this equation, we can have, $(u-15)(u-5)=0$
Therefore, $u=15$ and $u=5$ .
As, the value of $v$ , i.e., the speed of the car at the second point is $15m/s$ ,
So, $u$ = the speed of the car at the first point $=5m/s$.

Additional information:
If a particle is moving in a straight line, in a three-dimensional space, with constant acceleration, the three equations of motion can be applied to them to calculate the position, or speed, or acceleration of the particle. The three equations of motion are:
$v=u+at$
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
${{v}^{2}}={{u}^{2}}+2as$
Where
$u$ is the initial speed of the particle
$v$ is the final velocity of the particle,
$s$ is the displacement of the particle,
$a$ is the acceleration of the particle (in case of the bodies moving under the influence of gravity, the standard gravity $g$ is used)
And, $t$ is the time interval.

Note: These three equations of motion are often referred to as the SUVAT equations, where the word “SUVAT” is an acronym from the variables, $s$= displacement, $u$ = initial speed, $v$ = final speed, $a$ = acceleration, and $t$ = time.