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 = [tex]-\frac{18^{2} }{2*291.9}[/tex] = - [tex]\frac{324}{583.8}[/tex] = - 0.56 m/s 2
note: the answer is in negative because the Locomotive is decelerating and not accelerating
The magnitude of the minimum deceleration needed to avoid the accident is –0.55 m/s²
To solve the question given above, we'll begin by calculating the distance travelled during the reaction time. This can be obtained as follow:
Speed = 18 m/s
Time = 0.45 s
Speed = distance / time
18 = distance / 0.45
Cross multiply
Distance = 18 × 0.45
Thus, the engineer travelled a distance of 8.1 m during the reaction time.
Next, we shall the distance between the engineer and the car. This can be obtained as follow:
Total distance = 300 m
Distance during the reaction time = 8.1 m
= 300 – 8.1
Finally, we shall determine the magnitude of the deceleration needed to avoid the accident. This can be obtained as follow:
Initial velocity (u) = 18 m/s
Final velocity (v) = 0 m/s
Distance (s) = 291.9 m
0² = 18² + (2 × a × 291.9)
0 = 324 + 583.8a
Collect like terms
0 – 324 = 583.8a
–324 = 583.8a
Divide both side by 583.8
a = –324 / 583.8
Therefore, the magnitude of the deceleration needed to avoid the accident is –0.55 m/s²
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