Question

The speed of a cyclist reduces uniformly from $2.5m/s$ to $1.0m/s$ in $12s$ .
(A) Calculate the deceleration of the cyclist
(B) Calculate the distance travelled by the cyclist in this time

Answer 1

For solving this question, we need to use the two kinematic equations of motion. The first equation should have only acceleration as unknown and the second equation should have only displacement as unknown.

Formula used: The formulas used in solving this question are:
$v = u + at$
${v^2} - {u^2} = 2as$
$v =$ final velocity, $u =$ initial velocity, $s =$ displacement, $a =$ acceleration, and $t =$ time.

Complete step by step solution:
(A) Let $a$ be the deceleration of the cyclist.
According to the first equation of motion, we have
$v = u + at$
According to the question
$u = 2.5m/s$ , $v = 1.0m/s$ and $t = 12s$
Putting these in the above equation
$1 = 2.5 + a(12)$
$12a = 1 - 2.5$
Dividing by $12$ , we get
$a = \dfrac{{1 - 2.5}}{{12}}$
$a = - \dfrac{{1.5}}{{12}}$
Finally, we get
$a = - 0.125$
Hence, the deceleration of the cyclist is $0.125m/{s^2}$
(B) Let $d$ be the distance of the cyclist. As both the initial and final velocities of the cyclist are positive, so the motion of the object is unidirectional. Hence, the displacement covered by the cyclist equals its displacement.
According to the third equation of motion, we have
${v^2} - {u^2} = 2as$
Substituting
$u = 2.5m/s$ , $v = 1.0m/s$ and $a = - 0.125m/{s^2}$
${1^2} - {2.5^2} = 2( - 0.125)d$
On simplifying, we get
$1 - 6.25 = 2( - 0.125)d$
$- 5.25 = - 0.25d$
On rearranging
$0.25d = 5.25$
$d = \dfrac{{5.25}}{{0.25}}$
Finally, we get
$d = 21m$
Hence, the distance covered by the cyclist is $21m$ .

Note:
Always remember that the kinematic equations involve the displacement, not the distance covered by the particle. In this question, the motion was unidirectional, so the displacement could be replaced by the distance. In the case of bidirectional motion, the problem has to be divided into unidirectional motions and the corresponding displacements need to be calculated. Then the magnitudes of these displacements are needed to be added, for calculating the distance covered. It is a common mistake to consider the scalar quantities in the equation of motion.