1. If a particle's position is given by the expression x(t) = 5t³ - 52t meters, what is the acceleration of the particle after t = 5 seconds?

2. Given D = 5(t⁶ + 1)⁴ meters, find velocity (V) and acceleration (a) at t = 1 second.

3. An automobile moving at a constant velocity of 30 m/s passes a gasoline station. Two seconds later, another automobile leaves the gasoline station and accelerates at a constant rate of 2 m/s². How soon will the second automobile overtake the first?

To find acceleration of a particle, the function x ( t ) was differentiated twice, yielding 150 m/s 2 at t = 5 . For D = 5 ( t 6 + 1 ) 4 , the velocity was calculated as 960 m/s at t = 1 . Lastly, after solving for when the second automobile, accelerating at 2 m / s 2 , overtakes the first moving at 30 m / s , we determined the time from the passage of the first automobile.
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Answered by Anonymous | 2025-07-03

Position expression
take first derivative to get velocity expression : 15 t^2 - 52
take next derivative to get acceleration expression : a = 30t
so at t= 5 s acceleration is a = 30(t) = 30(5**) = 150 m/s^2 ** ;

Answered by jsimpson11000 | 2025-07-03