Question

Prove that v=ut+1/2 at2

Answer 1

Answer:

Consider a velocity-time graph where the object is moving at constant acceleration



(where uu denotes initial velocity and vv denotes final velocity)

The area of a velocity-time graph gives the displacement, therefore:

ss=Area of AOCD+Area of ADB=ut+12×t×(v−u)=ut+12×t×at∵v=u+at=ut+12at2s=Area of AOCD+Area of ADB=ut+12×t×(v−u)=ut+12×t×at∵v=u+ats=ut+12at2

Method 2 (uses calculus)

Displacement is the integral of velocity with respect to time:

s=∫vdts=∫vdt

Substitute v=u+atv=u+at into the integral:

ss=∫(u+at)dt=∫udt+∫atdt=ut+a∫tdt=ut+12at2+Cs=∫(u+at)dt=∫udt+∫atdt=ut+a∫tdts=ut+12at2+C

When t=0t=0 , s=0∴C=0s=0∴C=0 , so our equation reduces to

s=ut+12at

Answer 2

Answer:

acceleration is defined as change in velocity per unit time.

=> (v - u) / (t - 0) = a => v = u + at

Distance traveled is = average velocity * time duration

S= initial + final velocity /2 * duration

S = u+ v/ 2 * At

S = 1/2 ( u+ u+ at*t

S = 2ut + at2 / 2

S = it + 1/2 at2

Explanation:

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