A train moving at 72 km/h applies brakes and comes to a stop in 10 seconds. What is the deceleration?
Practice Questions
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Q1
A train moving at 72 km/h applies brakes and comes to a stop in 10 seconds. What is the deceleration?
2 m/s²
3 m/s²
4 m/s²
5 m/s²
First convert speed to m/s: 72 km/h = 20 m/s. Using a = (v - u)/t, we have a = (0 - 20)/10 = -2 m/s², so deceleration = 2 m/s².
Questions & Step-by-step Solutions
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Q
Q: A train moving at 72 km/h applies brakes and comes to a stop in 10 seconds. What is the deceleration?
Solution: First convert speed to m/s: 72 km/h = 20 m/s. Using a = (v - u)/t, we have a = (0 - 20)/10 = -2 m/s², so deceleration = 2 m/s².
Steps: 6
Step 1: Convert the speed from kilometers per hour (km/h) to meters per second (m/s). Use the conversion factor: 1 km/h = 1/3.6 m/s. So, 72 km/h = 72 / 3.6 = 20 m/s.
Step 2: Identify the initial speed (u) and final speed (v). The initial speed (u) is 20 m/s and the final speed (v) is 0 m/s because the train stops.
Step 3: Use the formula for acceleration (a): a = (v - u) / t, where t is the time taken to stop. Here, t = 10 seconds.
Step 4: Substitute the values into the formula: a = (0 - 20) / 10.
Step 5: Calculate the result: a = -20 / 10 = -2 m/s².
Step 6: Since deceleration is the magnitude of acceleration when slowing down, the deceleration is 2 m/s².
