The acceleration of a particle as it moves along a straight line is given by a = (4t – 1) m/ , 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.

## Your answer

 Your name to display (optional): Email me at this address if my answer is selected or commented on: 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)

1 answer
2 answers
1 answer
1 answer
1 answer
1 answer
1 answer
1 answer
2 answers
1 answer