Question

Ifs = 2t³ – 5t² + 4t – 3 (s= distance, t= time), find 

(i) The time when the acceleration is 14 ft/sec². 

(ii) The velocity and displacement at that time.

Answer 1

\(s = 2t^3 - 5t^2 + 4t - 3\)

\(\frac {ds}{dt} = \) Velocity

\(\frac {d^2s}{dt^2} = \) Acceleration

\(\frac {ds}{dt} = 6t^2 - 10t + 4\)

\(\frac {d^2s}{dt^2} = 12t - 10\)

(i) Acceleration = 14

⇒ 12t - 10 = 14

⇒ 12t = 24

⇒ t = \(\frac{24}{12}\) = 2 sec

(ii) \(\frac {ds}{dt} = 6t^2 - 10t + 4\)

\(\frac {ds}{dt}|_{t = 2} = 6(2)^2 - 10\times 2 + 4\)

\(= 24 - 20 +4\)

= 8 ft/sec

\(s = 2t^2 - 5t^2 + 4t -3\)

\(s|_{t = 2} = 16 -20 + 8 - 3\)

= 1 ft

Answer 2

If s = 2t³ – 5t² + 4t – 3

\(S(t= 2 sec) = 2\times2-5\times 2^2+4\times2-3\)

\(=16-20+8-3\)

\(=8-7 =1\,ft.\)

Comments

Thanks

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