Question:
A car moving at \(20\ \text{m/s}\) comes to rest with a uniform retardation of \(5\ \text{m/s}^2\). The stopping distance is:
A car moving at \(20\ \text{m/s}\) comes to rest with a uniform retardation of \(5\ \text{m/s}^2\). The stopping distance is:
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For stopping distance, final velocity is \(0\), and retardation is taken as negative acceleration.
Updated On: Jun 3, 2026
- \(20\ \text{m}\)
- \(30\ \text{m}\)
- \(40\ \text{m}\)
- \(50\ \text{m}\)
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The Correct Option is C
Solution and Explanation
Concept:
For uniformly accelerated or retarded motion, we use: \(\displaystyle v^2=u^2+2as\) where, \(u=\) initial velocity
\(v=\) final velocity
\(a=\) acceleration
\(s=\) distance
Step 1: Write the given values.
Initial velocity: \(\displaystyle u=20\ \text{m/s}\) Final velocity: \(\displaystyle v=0\ \text{m/s}\) Retardation: \(\displaystyle a=-5\ \text{m/s}^2\)
Step 2: Apply the formula.
\(\displaystyle v^2=u^2+2as\) Substitute the values: \(\displaystyle 0^2=20^2+2(-5)s\) \(\displaystyle 0=400-10s\) \(\displaystyle 10s=400\) \(\displaystyle s=40\ \text{m}\)
Step 3: Final conclusion.
Hence, the stopping distance is: \(\displaystyle \boxed{40\ \text{m}}\)
For uniformly accelerated or retarded motion, we use: \(\displaystyle v^2=u^2+2as\) where, \(u=\) initial velocity
\(v=\) final velocity
\(a=\) acceleration
\(s=\) distance
Step 1: Write the given values.
Initial velocity: \(\displaystyle u=20\ \text{m/s}\) Final velocity: \(\displaystyle v=0\ \text{m/s}\) Retardation: \(\displaystyle a=-5\ \text{m/s}^2\)
Step 2: Apply the formula.
\(\displaystyle v^2=u^2+2as\) Substitute the values: \(\displaystyle 0^2=20^2+2(-5)s\) \(\displaystyle 0=400-10s\) \(\displaystyle 10s=400\) \(\displaystyle s=40\ \text{m}\)
Step 3: Final conclusion.
Hence, the stopping distance is: \(\displaystyle \boxed{40\ \text{m}}\)
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