A PARTICLE IS MOVING ALONG A STRAIGHT LINE. ITS POSITION X AT TIME T IS GIVEN BY X=2T^3-3T^+2 FIND THE DISTANCE TRAVELLED BY THE PARTICLE IN THE FIRST 5 SECONDS
Question
A PARTICLE IS MOVING ALONG A STRAIGHT LINE. ITS POSITION X AT TIME T IS GIVEN BY X=2T^3-3T^+2 FIND THE DISTANCE TRAVELLED BY THE PARTICLE IN THE FIRST 5 SECONDS
Solution
To find the distance travelled by the particle in the first 5 seconds, we need to find the displacement of the particle at t=5 seconds and t=0 seconds and subtract the two.
The position function given is x = 2t^3 - 3t^2 + 2.
First, let's find the position of the particle at t=0 seconds:
x(0) = 2(0)^3 - 3(0)^2 + 2 = 2.
Next, let's find the position of the particle at t=5 seconds:
x(5) = 2(5)^3 - 3(5)^2 + 2 = 2(125) - 3(25) + 2 = 250 - 75 + 2 = 177.
The distance travelled by the particle in the first 5 seconds is the absolute difference between these two positions:
Distance = |x(5) - x(0)| = |177 - 2| = 175 units.
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