Question

A particle is projected in a straight line from a point O with a speed of $6ms^-^1$. At time t sec later its acceleration is $(1+2t)ms^-^2$ . For the time when t =4 , calculate for the particle :
1. its velocity
2. its distance from O

P.S : This question should be solved using integration.

Answer 1

Answer:

1. 26 m/s
2. 53 1/3 m

Step-by-step explanation:

The speed will be the initial speed plus the integral of acceleration:

  $\displaystyle v(t) = 6+\int_0^t{(1+2t)}\,dt=6+t+t^2\\\\v(t)=(t+1)t+6$

The position will be the integral of speed:

  $\displaystyle p(t)=\int_0^t{v(t)}\,dt=6t+\dfrac{1}{2}t^2+\dfrac{1}{3}t^3\\\\p(t)=\dfrac{((2t+3)t+36)t}{6}$

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1. The velocity at t=4 is v(4) = (4+1)4 +6 = 26 m/s

2. The position at t=4 = ((2·4 +3)4 +36)4/6 = (44 +36)(2/3) = 160/3 = 53 1/3 m

At t=4, the particle will be 53 1/3 m from O and will be moving at a speed of 26 m/s.