The acceleration of a particle as it moves along a straight line is given by a = (4t – 1) m/image, where t is in seconds. If s = 2 m and v = 5 m/s when t = 0, determine the particle’s velocity and position when t = 7s. Also, determine the total distance the particle travels during this time period.

in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Best answer

a=dv/dt=4t-1.

Integrating:

v(t)=2t²-t+c₁, v(0)=c₁=5m/s, v(t)=2t²-t+5.

v=ds/dt=2t²-t+5.

Integrating:

s(t)=2t³/3-t²/2+5t+c₂, s(0)=c₂=2m, s(t)=2t³/3-t²/2+5t+2.

s(7)=686/3-49/2+35+2=1447/6 m.

Distance travelled from t=0 (when s=2m) to t=7=1435/6 m.

This can also be calculated from the definite integral:

∫₀⁷(2t²-t+5)dt=[2t³/3-t²/2+5t]₀⁷=(686/3-49/2+35)=1435/6 m.

(Note that at t=0, a=-1m/s², v=5m/s, s=2m; and at t=7, a=27m/s², v=96m/s, s=1447/6 m.)

by Top Rated User (839k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
86,303 questions
92,362 answers
2,251 comments
23,927 users