Question

A train is moving with a velocity of 72 km/h. The brakes are applied to produce negative acceleration and stop the the motion of the train uniformly. If the train is stopped after 50 m away from the place where brakes were applied, find the negative acceleration of the train.

Answer 1

Hey There,


Here, we have the following data:

Initial Velocity = u = 72 km/hr = $72 \times \frac{5}{18}$ m/s = 20 m/s

Final Velocity = v = 0 m/s

Distance Covered = s = 50 m


Let the acceleration value be a.

From the equations of motion, we have:

$v^2 = u^2 + 2as \\ \\ \\ \implies 0 = (20)^2 + 2a(50) \\ \\ \\ \implies -100a = 400 \\ \\ \\ \implies \boxed{a = -4 \, \, m / s^2}$


Thus, the acceleration is - 4 $\textbf{m/ s^2 }$



Hope it helps
Purva
Brainly Community

Answer 2

Hey !!!!! Your answer =>

# Physics # -- Numericals using the equation of motion.

We are provided with some things in it that are -

$distance = 50 \: m \\ \\ initial \: velocity = u = 72 \: \frac{km}{hr} \\ \\ to \: change \: it \: in \: \frac{m}{s} multiply \: it \: with \: \frac{5}{18} \\ \\ u \: = 72 \times \frac{5}{18} \\ \\ u = 20 \: \frac{m}{s} \\ \\ final \: velocity \: = v = 0 \: \frac{m}{s} \\ \\ this \: is \: coz \: the \: body \: was \: brought \: at \: rest. \\ \\ so \: lets \: do \: it \: by \: the \: third \: equation \: of \: motion \: that \: is \: \\ \\ {v}^{2} = {u}^{2} + 2as \\ \\ {0}^{2} = {20}^{2} + 2 \times a \times 50 \\ \\ 0 = 400 + 100a \\ \\ - 400 = 100a \\ \\ \frac{ - 400}{100} = a \\ \\ a = - 4 \: \frac{m}{ {s}^{2} }$


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