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#stopping distance

6 public questions tagged with this topic.

A car moving at 50m/s decelerates uniformly to rest over 125m. What is the time taken to stop?

Use v2=v02+2ax to find a: 0=(50)2+2a(125)⇒0=2500+250a⇒a=−10m/s2. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 5 s as the result, so option D is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Uniform Acceleration and Graphical Analysis

A car moving at 20m/s decelerates uniformly and stops after traveling 50m. What is the deceleration?

Use v2=v02+2ax. Here, v=0, v0=20m/s, x=50m. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 4 m/s² as the result, so option B is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations

A car moving at 35m/s decelerates uniformly to rest over 87.5m. What is the deceleration?

Use v2=v02+2ax. Here, v=0, v0=35m/s, x=87.5m. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 7 m/s² as the result, so option B is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations

A car traveling at 54km/h comes to rest in 10s with uniform deceleration. What is the magnitude of the deceleration?

Convert speed: 54km/h=54⋅10003600=15m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 1.5 m/s² as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Kinematic Equations and Uniformly Accelerated Motion