दिए गए चित्र में एक कण की गति का त्वरण-समय आरेख दिखाया गया है। यदि कण का प्रारंभिक वेग 8 m/s है, तो 16वें सेकंड के अंत तक कण का विस्थापन कितना है?

Concept:

The area under the acceleration-time graph represents the change in velocity.

∴ The area under acceleration - time graph = ΔV

Similarly, the area under the velocity-time graph represents the total displacement. 

∴ Area under velocity - time graph = displacement (S)

Calculation:

Given:

In the given problem, acceleration decreases to zero and then increases linearly.

Now we know, \(dV = a.dt\)

\(dS = V. dt = (u + dV) . dt\) ; where u is the initial velocity. 

Here u = 8 m/s given.

The above problem can be divided into two portions:

(1) t = 0 s to t = 8 s

The line equation of the a-t curve can be written as \(a = {8-t \over 8}\)

∴ \(dV = {8-t \over 8} . dt\)

∴ \(∫_{0}^{V}dV = ∫_{0}^{t}{8-t \over 8} . dt\)

∴ \(V = {8t-{t^2 \over2} \over 8}\)

Now, \(dS = V . dt\)

∴ \(S1 = ∫_{0}^{8}(u + {8t-{t^2 \over2} \over 8}). dt \;\)

After integrating and substitution of values,

S₁ = 85.33 m

∴ distance travelled by a particle in t = 0 to t = 8 s is 85.33 m.

Similarly area under the curve (A₁) will be = area of triangle covered by the line = 1/2 * 1 * 8 = 4

(1) t = 8 s to t = 16 s

The line equation of the a-t curve can be written as \(a = {t-8 \over 4}\)

The initial velocity for this section will be the velocity at t = 8 s

\(V_{t\ = \ 8} \) = area (A₁) + initial velocity at t = 0, s = 4 + 8 = 12 m/s

∴ \(dV = {t-8 \over 4} . dt\)

∴ \(∫_{12}^{V}dV = ∫_{8}^{t}{t-8 \over 4} . dt\;\)

∴ \(V = {t^2 \over 8}-2t+20\)

Now displacement covered  S₂ = ∫ V dt = \(\int_{8}^{16}({t^2 \over 8}-2t+20) dt\)

∴ after integration and substituting values S₂ = 117.33 m

Total distance,

S = S₁ + S₂ = 85.33 + 117.33 = 202.67 m