A732 kg car stopped at an intersection is rear-ended by a 1720 kg truck moving with a speed of 15.5m/s. if the car was in neutral and its breaks were off, so the collisions are approximately elastic, find the final speed of both vehicles after the collisions.

The final velocity of the car is 6.22 m/s and the final velocity of the truck is 21.72 m/s.

The given parameters;

The final velocity of both vehicles is calculated by applying the principle of conservation of linear momentum;

Apply one-directional linear velocity equation;

substitute the value of v₁ into the above equation;

Thus, the final velocity of the car is 6.22 m/s and the final velocity of the truck is 21.72 m/s.

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Truck= 6.25 m/s

Car= 21.7 m/s

Explanation:

For elastic collision, the truck's final velocity is given by

The car's final velocity is given by

Where m and u denote masses and velocity respectively, subscripts 1 and 2 denote truck and car respectively.

Substituting 1720 Kg for mass of truck, 732 for mass of car and initial velocity as 15.5 m/s

Therefore, the final velocity of truck is 6.25 m/s while the car's final velocity is 21.7 m/s

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.