Stopping distance and time under uniform deceleration: A car moving at 20 m/s is brought to rest uniformly within a distance of 40 m. How long does it take to stop?

Introduction / Context:
Uniformly accelerated (or decelerated) rectilinear motion is governed by constant-acceleration kinematics. Stopping a vehicle in a given distance from a known initial speed is a standard application of these formulas.

Given Data / Assumptions:

• Initial speed u = 20 m/s.
• Final speed v = 0 m/s (comes to rest).
• Stopping distance s = 40 m.
• Acceleration a is constant (uniform deceleration).

Concept / Approach:

Use v^2 = u^2 + 2 a s to find the constant acceleration a. Then use v = u + a t to compute the stopping time t. Signs indicate deceleration (a negative value), but magnitudes are sufficient to get time.

Step-by-Step Solution:

Verification / Alternative check:

Average speed during uniform deceleration is (u + v)/2 = 10 m/s. Time = distance / average speed = 40 / 10 = 4 s, confirming the result.

Why Other Options Are Wrong:

2 s and 3 s imply much larger decelerations; 5 s and 6 s imply smaller decelerations insufficient to stop within 40 m.

Common Pitfalls:

Sign errors in a; forgetting that average speed equals mid-value when acceleration is constant; mixing units.

Final Answer:

4 s