- 2068 Shrawan
Particle with x = t³/3 − 6t² − 15t
Short answer
Velocity zero at t=13.14 s. Position -476.8 m, distance 476.8 m, acceleration 14.28 m/s².
The position of a particle which moves along a straight line is defined by the relation x = t³/3 − 6t² − 15t, where x is in metre and t in seconds. Determine: i) The time at which velocity will be zero. ii) The position and distance travelled by the particle at that time. iii) The acceleration of the particle at that time.
Velocity and acceleration
x = t³/3 − 6t² − 15t
- v = t² − 12t − 15 (m/s)
- a = 2t − 12 (m/s²)
(i) Time when v = 0: t² − 12t − 15 = 0, so t = 6 + √51 = 13.14 s (the negative root is rejected).
(ii) Position and distance travelled at that time x = 13.14³/3 − 6(13.14)² − 15(13.14) = 756.5 − 1036.1 − 197.1 = −476.8 m At t = 0, v = −15 m/s. The velocity first becomes more negative, reaching its minimum of −51 m/s at t = 6 s (where a = 0), and then rises to zero at 13.14 s. It stays negative throughout, so the particle moves only in the negative direction and the distance travelled is 476.8 m.
(iii) Acceleration at that time: a = 2(13.14) − 12 = 14.28 m/s²
✓ Answer checked by Claude Opus 5.5 (high effort)
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