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Equations of Motion and Problem Solving

Problems and solutions using the standard equations of motion to analyze and solve motion-related questions in physics.

28 questions

A particle’s position is given by x\=2t+t2 and y\=3t−2t2 (in meters and seconds). What is its speed at t\=2s?

Velocity: vx=dxdt=2+2t,vy=dydt=3−4t. 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 6 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving

A particle’s position is given by x\=3t2−4t and y\=5t (in meters and seconds). What is its speed at t\=1s?

Velocity: vx=dxdt=6t−4,vy=dydt=5. 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/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving

A stone tied to a string moves in a circle of radius 1.5m with a frequency of 2Hz. What is its centripetal acceleration?

Angular speed ω=2πf=2π×2=4πrad/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 150 m/s² as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving

A stone is projected horizontally with a speed of 10 m/s from a height of 20 m. How long does it take to reach the groun

For vertical motion: y = v₀y t + (1/2) a\_y t², where v₀y = 0, a\_y = -g. 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 2 s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving

A plane flies east at 40m/s while a wind blows north at 30m/s. What is the magnitude of the plane’s velocity relative to

Velocity components: vx=40m/s,vy=30m/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 50 m/s as the result, so option B is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving

A ball is thrown at 30 m/s at 60° to the horizontal. What is the time of flight? (Take g = 10 m/s²)

Time of flight T\_f = (2 v₀ sin θ₀) / g. 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 3 s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving

A ball is thrown vertically upwards with an initial speed of 28m/s. What is the total time of flight? (Take g\=10m/s2)

Time to max height: v=v0+at, 0=28−10t⇒t=2.8s. 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 2.8 s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving

A train moving at 90km/h accelerates at 0.8m/s2 for 15s, then decelerates at 2.5m/s2 to rest. What is the total distance

Speed: 90km/h=25m/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 730 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving