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A particle travels in a straight line from A to B in 20s. Its acceleration t seconds after leaving A is ms^(-2), where a=(3)/(160)t^(2)-(1)/(800)t^(3). It is given that the particle comes to rest at B.

(i) Show that the initial speed of the particle is zero.
(ii) Find the maximum speed of the particle.
(iii) Find the distance AB​

A particle travels in a straight line from A to B in 20s. Its acceleration t seconds-example-1
User Mehul Chauhan
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1 Answer

22 votes
22 votes

Answer:

  • initial speed is final speed: 0
  • maximum speed is about 5.273 m/s
  • distance A to B is 50 meters

Explanation:

You have a particle whose acceleration is a(t) = 3t²/160 -t³/800 and whose velocity at t=20 is 0. You want to show the initial speed is 0, find the maximum speed, and the distance traveled in 20 seconds.

(i) Initial speed

The speed is the sum of the initial speed and that due to acceleration. The speed due to acceleration is the integral of acceleration:


\displaystyle s(20)=s_0+\int_0^(20){(1)/(800)(15t^2-t^3)\,dt}}=s_0+\left((t^3)/(3200)(20-t)\right)_0^(20)=s_0=0

The change in speed over 20 seconds is zero, so the final speed is the initial speed. The final speed is zero, so the initial speed is 0.

(ii) Maximum speed

The maximum speed is where the acceleration is zero.

a(t) = 0 = (t²/800)(15 -t)

The acceleration is zero at t=0 and at t=15.

From the above integral, the maximum speed is ...

s(15) = 15³(20 -15)/3200 = 5.2734375 m/s ≈ 5.27 m/s

(iii) Distance

The distance traveled is the integral of the speed.


\displaystyle d=(1)/(3200)\int_0^(20){(20t^3-t^4)}\,dt=\left.(t^4(25-t))/(16000)\right|_0^(20)}=50

The particle traveled 50 meters from A to B.

A particle travels in a straight line from A to B in 20s. Its acceleration t seconds-example-1
User Mdeous
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3.1k points