Question

a train named iris starts from rest and accelerates uniformly until it has traveled 2.1 km and acquired a forward velocity of 24 m/s. the train then moves at a constant velocity of 24 m/s for 400 s. the train then slows down uniformly at 0.065 m/s² until it is brought to a halt. the distance traveled by the train while slowing down is closest to .........3.8 km.4.0 km.4.2 km.4.4 km.this is a required question

Explanation:

Step1: Identify known variables

Initial velocity $u = 24\ \text{m/s}$, final velocity $v = 0\ \text{m/s}$, acceleration $a = -0.065\ \text{m/s}^2$

Step2: Select kinematic equation

Use $v^2 = u^2 + 2as$ to solve for distance $s$

Step3: Rearrange for distance $s$

$s = \frac{v^2 - u^2}{2a}$

Step4: Substitute values

$s = \frac{0^2 - (24)^2}{2\times(-0.065)} = \frac{-576}{-0.13} \approx 4430.77\ \text{m}$

Step5: Convert to kilometers

$s = \frac{4430.77}{1000} \approx 4.4\ \text{km}$

Answer:

4.4 km.