Question

An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first sees the car, the locomotive is 260 m from the crossing and its speed is 26 m/s. If the engineer's reaction time is 0.51 s, what should be the magnitude of the minimum deceleration to avoid an accident? Answer in units of m/s 2

Answer 1

Answer:

The right answer is "1.369 m/s²".

Explanation:

The given values are:

Distance (s)

= 260 m

Initial speed (u)

= 26 m/s

Reaction time (t')

= 0.51 s

During reaction time, the distance travelled by locomotive will be:

⇒  $s'=ut'$

        $=26\times 0.51$

        $=13.26 \ m$

Remained distance between locomotive and car:

⇒  $x=s-s'$

         $=260-13.26$

         $=246.74 \ m$

Now,

The final velocity to avoid collection is, V = 0 m/s

From third equation of motion:

⇒  $V^2=u^2+2ax$

On putting the estimated values, we get

⇒  $0=(26)^2+2\times a\times 246.74$

⇒  $0=676+493.48a$

⇒  $493.48a=-676$

⇒            $a=-\frac{676}{493.48}$

⇒            $a=1.369 \ m/s^2$