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A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t2 – 4t + 5 cm/second. What will be the distance travelled by the particle during first 3 seconds after the start?
- a)21 cm
- b)22 cm
- c)23 cm
- d)24 cm
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A particle moves in a straight line and its velocity v at time t secon...
Let, x be the distance travelled by the particle in time t seconds.
Then, v = dx/dt = 3t2 – 4t + 5
Or ∫dx = ∫ (3t2 – 4t + 5)dt
So, on integrating the above equation, we get,
x = t3 – 2t2 + 5t + c where, c is a constant. ……….(1)
Therefore, the distance travelled by the particle at the end of 3 seconds,
= [x]t = 3 – [x]t = 0
= (33 – 2*32 + 5*3 + c) – c [using (1)]
= 24 cm.
Then, v = dx/dt = 3t2 – 4t + 5
Or ∫dx = ∫ (3t2 – 4t + 5)dt
So, on integrating the above equation, we get,
x = t3 – 2t2 + 5t + c where, c is a constant. ……….(1)
Therefore, the distance travelled by the particle at the end of 3 seconds,
= [x]t = 3 – [x]t = 0
= (33 – 2*32 + 5*3 + c) – c [using (1)]
= 24 cm.
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