Displacement of a particle moving on a straight line is given by x=16t-2t2. Distance travelled by the particle during the first 2sec. isS1 and during first 6sec . is S2 find 3S1/S2?

Question

Answer

Displacement of a Particle

The displacement of a particle moving on a straight line is given by:

x = 16t - 2t²

Finding the Distance Travelled

The distance travelled by a particle during a given time period is the total length of the path travelled by the particle during that time period. To find the distance travelled by the particle during the first 2 seconds, we can use the following formula:

S₁ = ∫₀²|v(t)| dt

where v(t) is the velocity of the particle at time t. The absolute value of v(t) is taken because distance is always a positive quantity.

We can find the velocity of the particle by taking the derivative of the displacement equation:

v(t) = dx/dt = 16 - 4t

Substituting this into the formula for S₁, we get:

S₁ = ∫₀²|16 - 4t| dt = ∫₀²(16 - 4t) dt = 16t - 2t² from t=0 to t=2

Therefore, S₁ = 20 units.

To find the distance travelled by the particle during the first 6 seconds, we can use the same formula:

S₂ = ∫₀⁶|v(t)| dt

However, we need to split this integral into two parts because the velocity of the particle changes direction at t=4 seconds. Therefore, we have:

S₂ = ∫₀⁴|16 - 4t| dt + ∫₄⁶|16 - 4t| dt

Solving these integrals, we get:

S₂ = (16t - 2t²) from t=0 to t=4 + (-16t + 2t² - 96) from t=4 to t=6

Therefore, S₂ = 64 units.

Finding 3S₁/S₂

Now that we have found S₁ and S₂, we can use them to find 3S₁/S₂:

3S₁/S₂ = 3(20)/64 = 15/8

Explanation

Answer

X(2sec)=16(2)-(2)^2=28m with velocity 1=8m/s
distance is the same as displacement.
X(6sec)=16(6)-(6)^2=60m but with velocity opposite to V1. the distance become twice of the displacement. S2=2(X(6sec)=120m
3S1/S2=3(28)m/120m=0.7

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