Distance travelled by the particle on x-axis
Question
A particle moves on x-axis as per equation \( x = t^3 - 9t^2 + 15t + 2 \) m. Distance travelled by the particle between \( t = 0 \) and \( t = 5 \) s is (A) 25 m (B) 39 m (C) 23 m (D) 52 mAnswer
Given position function: \( x(t) = t^3 - 9t^2 + 15t + 2 \)
Step 1: Find velocity \( v(t) = \frac{dx}{dt} \)
\[ v(t) = 3t^2 - 18t + 15 \]
Step 2: Find times when velocity is zero (turning points) to find change in direction:
\[ 3t^2 - 18t + 15 = 0 \implies t^2 - 6t + 5 = 0 \implies (t-5)(t-1) = 0 \]
So, \( t = 1 \) s and \( t = 5 \) s are points where velocity changes sign.
Step 3: Calculate position at \( t=0, 1, 5 \):
- \( x(0) = 0 - 0 + 0 + 2 = 2 \)
- \( x(1) = 1 - 9 + 15 + 2 = 9 \)
- \( x(5) = 125 - 225 + 75 + 2 = -23 \)
Step 4: Distance travelled is sum of absolute displacements between turning points:
\[ \text{Distance} = |x(1) - x(0)| + |x(5) - x(1)| = |9 - 2| + |-23 - 9| = 7 + 32 = 39 \text{ m} \]
Final Answer
\boxed{39 \text{ m}}\end{div}