Question

$\huge\bf\red{Question}$
The acceleration of a particle travelling along a straight line is a = kt^2, where k is constant. if at t = 0, v = 0 then speed of particle at t = 3 sec will be

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Answer 1

Answer:

The position of particle is given as x=37+27t−t3

Thus the velocity of particle  v=dtdx=27−3t2

According to problem,

v=0⇒27−3t2=0

Here, we get t=327=9=3s

Thus the distance of particle when it comes to rest   x(t=3 s)=37+27×3−(3)3=91 m

Explanation:

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Answer 2

Correct Question :-

The acceleration of a particle travelling along a straight line is a = kt² , where k is constant ( k = 1 ). If at t = 0, v = 0 then speed of particle at t = 3 sec will be

Answer :-

We are given the equation of acceleration with respect to time. So by integrating we can get the equation of velocity with respect to time.

➩ $\sf \displaystyle\sf\int adt = \Delta v$

➩ $\sf \displaystyle\sf v = \int kt^2 dt$

➩ $\sf v = \dfrac{kt^3}{3}$

So now we get the equation of velocity with respect to time.

Now we are asked velocity at t = 3 sec

Substituting the value of t :-

➩ $\sf v = \dfrac{kt^3}{3}$

➩ $\sf v = \dfrac{1 \times 3^3}{3}$

➩ $\sf v = 3^2$

➩ $\sf v = 9 m/s$

Velocity at t = 3 sec will be 9 m/s