(✰✰✰✰) A particle moves along a line. At time t (in seconds), the particle's velocity is v(t) = -t² + 5t − 6 (in m/s). Find the particle's net displacement and total distance traveled over the time interval [0, 5].
(✰✰✰✰) A particle moves along a line. At time t (in seconds), the particle's velocity is v(t) = -t² + 5t − 6 (in m/s). Find the particle's net displacement and total distance traveled over the time interval [0, 5].
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![(★★☆☆) A particle moves along a line. At time t (in seconds), the particle's velocity is v(t) = −t² + 5t − 6
(in m/s). Find the particle's net displacement and total distance traveled over the time interval [0, 5].
Net displacement = (-5^3/3+5/2(5^2)-6(5))-(-0^3/3+5/2(0^2) (m)
Total distance traveled = -14/3-1/6+4/3
(m)
To understand the terms "net displacement" and "total distance travelled," imagine a particle that moves
from x = 0 to x = 2, then back to x = 1. Its net displacement is 1, since it started at position 0 and ended
at position 1. Its total distance travelled is 3, since it moved 3 units all together.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49ece5d2-14dd-4033-91d1-76ec248e0625%2F61b3b671-ca6d-44bb-aaaa-515ad92eb1f5%2Fbb41jp_processed.png&w=3840&q=75)
Transcribed Image Text:(★★☆☆) A particle moves along a line. At time t (in seconds), the particle's velocity is v(t) = −t² + 5t − 6
(in m/s). Find the particle's net displacement and total distance traveled over the time interval [0, 5].
Net displacement = (-5^3/3+5/2(5^2)-6(5))-(-0^3/3+5/2(0^2) (m)
Total distance traveled = -14/3-1/6+4/3
(m)
To understand the terms "net displacement" and "total distance travelled," imagine a particle that moves
from x = 0 to x = 2, then back to x = 1. Its net displacement is 1, since it started at position 0 and ended
at position 1. Its total distance travelled is 3, since it moved 3 units all together.
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