A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.0 seconds, coasts for 2.4 s , and then slows down at a rate of 1.5m/s2 for the next stop sign.

d = 112.8m

the stop signs are 112.8m apart.

A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.0 seconds, coasts for 2.4 s , and then slows down at a rate of 1.5m/s2 for the next stop sign. How far apart are the stop signs?

Explanation:

The motion are in three stages;

1. It accelerates at 2.0 m/s2 for 6.0 seconds

2. It coasts for 2.4 s

3. then slows down at a rate of 1.5m/s2 for the next stop sign.

1. It accelerates at 2.0 m/s2 for 6.0 seconds;

distance d1 = ut + 0.5at^2

u = 0

t = 6.0 s

a = 2.0 m/s^2

d1 = 0 + 0.5×2.0(6^2)

d1 = 36 m

2. It coasts for 2.4 s ;

Final Speed v2 = at = 2 × 6

v2 = 12 m/s

Distance d2 = vt = v2(t2)

T2 = 2.4 s

Substituting the values;

d2 = 12×2.4 = 28.8 m

3. then slows down at a rate of 1.5m/s2 for the next stop sign.

Time taken to decelerate t3 = ∆v/a = 12/1.5 = 8 s

Distance d3 = ut + 0.5at^2

u = 12 m/s

a = -1.5 m/s^2

t = 8 s

d3 = 12(8) +0.5(-1.5 ×8^2)

d3 = 48 m

Total distance d = d1 +d2 +d3

d = 36m + 28.8m + 48m

d = 112.8m

the stop signs are 112.8m apart.

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.

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