Question

A body travels 200 cm in the first 2s and 220 cm in the next 5sec Calculate the velocity of the body at the end of 7sec from t he start assuming uniform acceleration through out the journey

Answer 1

I assume the acceleration is constant (otherwise there's no unique solution). 

Let V₀ = initial velocity; V₁ = velocity after 2 sec; V₂ = velocity after 7 sec. 

We know that average velocity is defined as: 

V_avg = distance / (elapsed time) 

In cases where acceleration is constant, it is also true that V_avg is the average of the initial and final speeds: 

V_avg = ½(Vi + Vf) (when acceleration is constant). 

That means, when acceleration is constant: 

½(Vi + Vf) = distance / (elapsed time) 

So let's apply that relation to the data in the question. 

"...200cm in the first 2 second..." 

That means: 

½(V₀+V₁) = 200cm/2sec = 100 cm/s 

"...220cm in the next 5 second." 

That means: 

½(V₁+V₂) = 220cm/5sec = 44 cm/s 

Now subtract the 2nd equation from the first, and you get: 

½(V₀−V₂) = 56 cm/s 

But we also know that it traveled (200cm+220cm)=420 cm, in (2s+5s)=7 seconds. So applying the "average speed" formula over the full 7 seconds we get: 

½(V₀+V₂) = 420cm/7s = 60 cm/s 

Now take this equation and subtract the one above it, and you get: 

V₂ = 4 cm/s

Answer 2

Distance =200 +220= 420 Time=7seconds Velocity=dis/time =420/7 =60cm/sec.