Question
A car traveling with a uniform velocity of 20 m/s is subjected to a uniform retardation of 2 m/s². How long will it take to come to rest, and what distance will it cover before stopping?
Quick Answer
Final velocity v = 0
Initial velocity u = 20 m/s
Retardation a = –2 m/s²
Time = (v – u) / a = (0 – 20) / –2 = 10 s
Distance = (u + v) / 2 × t = (20 + 0) / 2 × 10 = 100 m
The car will take 10 seconds to stop and will cover 100 metres before coming to rest.
Understanding the question with detailed explanations
The problem clearly states that the car is initially moving at 20 m/s. The braking introduces a uniform retardation of 2 m/s². Retardation is simply negative acceleration, meaning the car’s velocity decreases at a steady rate of 2 m/s every second until it becomes zero. The first part of the task is to calculate the time it takes for the velocity to reduce from 20 m/s to 0 m/s.
The second part requires us to find the stopping distance. Once the car begins to decelerate, it continues to travel some distance even though its velocity is decreasing. The distance covered during deceleration is not just speed multiplied by time, because the speed is changing every second. Instead, we use the average velocity method or the appropriate equation of motion.
A glimpse of the final answer
After applying the equations, we find that the car takes 10 seconds to stop. This is logical because if the velocity decreases by 2 m/s each second, then in 10 seconds, it will lose all of its 20 m/s.
The stopping distance comes out as 100 metres. This result is reasonable because while the car is decelerating, it still travels forward significantly due to its initial velocity. A car covering 100 metres while braking from 20 m/s aligns with real-world experiences, reinforcing that the calculation is correct.
Data
From the problem:
- Initial velocity, u = 20 m/s
- Final velocity, v = 0 m/s (since the car comes to rest)
- Retardation, a = –2 m/s²
What we need:
- Time, t = ?
- Distance, s = ?
All quantities are already in SI units, so no conversion is necessary.
Formula
Two equations of motion are most relevant here:
- v = u + at
- s = (u + v) / 2 × t
The first equation relates velocity, acceleration, and time. Since we know u, v, and a, we can directly solve for t. The second equation gives the displacement when motion has uniform acceleration or retardation, by taking the average velocity and multiplying it by time.
These formulas are perfectly suited because the acceleration is constant. If the retardation had been non-uniform, we would need calculus, but for this case, simple linear equations of motion are enough.
Solution (solving the problem)
Step 1: Time to come to rest
v = u + at
0 = 20 + (–2) × t
–20 = –2t
t = 10 seconds
Step 2: Distance covered before stopping
s = (u + v) / 2 × t
s = (20 + 0) / 2 × 10
s = 10 × 10
s = 100 metres
Final Answer
The car takes 10 seconds to come to rest under uniform retardation of 2 m/s².
During this time, it covers a stopping distance of 100 metres before halting completely.
Helpful Explanation
This example illustrates how braking distance depends on both initial speed and rate of deceleration. At 20 m/s (which is 72 km/h), the car takes a considerable distance of 100 metres to stop, even with steady braking. This shows why higher speeds can be dangerous, doubling the speed would more than double the stopping distance.
A common mistake students make is forgetting to apply the negative sign to retardation. If acceleration is taken as positive instead of negative, the time and distance calculations would go wrong. Another useful exam tip is to always check whether the problem asks for distance, displacement, or velocity, and to select the formula that directly connects the given data with the unknowns.