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Zorluk: ZorSpeed, Velocity, and Acceleration

A high-speed rail train is traveling at an initial velocity of 216 km/h216 \text{ km/h}. To safely navigate a sharp curve, the train must reduce its velocity to 108 km/h108 \text{ km/h}. If the train applies its braking system to produce a constant deceleration of 1.5 m/s21.5 \text{ m/s}^2, how many seconds will it take to reach the required speed?

Cevap: 20 seconds

Cevap

It will take 20 seconds for the train to reach the required speed.
The correct answer is found by first converting both velocities to meters per second to match the acceleration's units. 216 km/h216 \text{ km/h} is equivalent to 60 m/s60 \text{ m/s}, and 108 km/h108 \text{ km/h} is equivalent to 30 m/s30 \text{ m/s}. The change in velocity is −30 m/s-30 \text{ m/s}. Using the formula t=Δv/at = \Delta v / a, dividing −30 m/s-30 \text{ m/s} by the deceleration of −1.5 m/s2-1.5 \text{ m/s}^2 results in exactly 20 seconds20 \text{ seconds}.

Adım Adım Çözüm

1
Convert the initial velocity (viv_i) from km/h to m/s.
216 km/h÷3.6=60 m/s216 \text{ km/h} \div 3.6 = 60 \text{ m/s}
Standardizing units to match the acceleration given in m/s^2 is required before calculation.
2
Convert the final velocity (vfv_f) from km/h to m/s.
108 km/h÷3.6=30 m/s108 \text{ km/h} \div 3.6 = 30 \text{ m/s}
All velocity variables must be in meters per second to be compatible with the acceleration.
3
Calculate the change in velocity (Δv\Delta v).
Δv=30 m/s−60 m/s=−30 m/s\Delta v = 30 \text{ m/s} - 60 \text{ m/s} = -30 \text{ m/s}
Acceleration is the rate of change of velocity, so finding the total change is the necessary intermediate step.
4
Apply the acceleration formula a=Δvta = \frac{\Delta v}{t} to solve for time (tt).
t=−30 m/s−1.5 m/s2=20 secondst = \frac{-30 \text{ m/s}}{-1.5 \text{ m/s}^2} = 20 \text{ seconds}
Dividing the total change in velocity by the rate of deceleration yields the elapsed time.

Anahtar Kavram

Kinematics: Calculating time from constant acceleration and velocity change while managing unit conversions.
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