Question

A cyclist starting from rest moves with uniform acceleration and covers 120m in 10s. Then his acceleration in ms^-2 is

• A. 5
• B. 1.5
• C. 2.4
• D. 3
• E. 4.8

Answer 1

The Correct Option is C

We can use the kinematic equation to find the acceleration: \[ s = ut + \frac{1}{2} a t^2 \] Where: - \( s \) is the distance covered (120 m), - \( u \) is the initial velocity (0 m/s, since the cyclist starts from rest), - \( a \) is the acceleration, - \( t \) is the time taken (10 s). Substituting the known values into the equation: \[ 120 = 0 \times 10 + \frac{1}{2} a (10)^2 \] Simplifying: \[ 120 = \frac{1}{2} a \times 100 \] \[ 120 = 50a \] \[ a = \frac{120}{50} = 2.4 \, \text{m/s}^2 \]

The correct option is (C) : \(2.4\ \text{m/s}^2\)

Answer 2

The cyclist starts from rest, so initial velocity $u = 0$. He covers distance $s = 120$ m in time $t = 10$ s. Use the equation of motion:  

$s = ut + \frac{1}{2} a t^2$ 
Substituting values: $120 = 0 \cdot 10 + \frac{1}{2} a (10)^2$ 
$120 = \frac{1}{2} a \cdot 100$ 
$120 = 50a$ 
$a = \frac{120}{50} = 2.4 \ \text{m/s}^2$ 

Correct answer: 2.4