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 300 m from the crossing and its speed is 18 m/s. If the engineer’s reaction time is 0.45 s, what should be the magnitude of the minimum deceleration to avoid an accident? Answer in units of m/s 2 .

Answer 1

Answer:

a=0.555m/s^2

Explanation:

First we find the distance traveled from the moment the engineer reacts to the car, assuming uniform movement

X=VT

X=(18)(0.45)=8.1m

then we find the distance at which the deceleration begins, which is obtained by subtracting the total distance with the inner result

X=300-8.1=291.9

finally we use the equation for constant acceleration

Vf=0 final speed

Vo=18m/s= initial speed

X=291.9m

(Vf^2-Vo^2)/2X=a

(0-18^2)/(2*291.9)=a

a=0.555m/s^2

Answer 2

Answer:

- 0.56 m/s 2

Explanation:

To calculate the magnitude of deceleration to avoid the collision

distance = 300 m

initial speed = 18 m/s

reaction time ( t ) = 0.45 s

first calculate the distance traveled by the engineer/train during his reaction time

D₁ = initial speed * reaction time = 18 * 0.45 = 8.1 m

Then the remaining distance before impact between the car and Train would be

= distance - distance traveled during reaction time

= 300 m - 8.1 m = 291.9 m

from the equation of motion of objects

V² = U² + 2ax  (equation 1)

To avoid impact the final velocity( v )  would be ( 0 )

u = initial speed ( 18 m/s )

a = acceleration/deceleration ( ? )

x = remaining distance before collision ( 291.9 m )

hence equation 1 becomes

a = $-\frac{18^{2} }{2*291.9}$   =  - $\frac{324}{583.8}$  = - 0.56 m/s 2

note: the answer is in negative because the Locomotive is decelerating and not accelerating