Question

Тело движется равномерно со скоростью 3 м/с в течение 5 секунд, после чего получает ускорение 0,2 м/с^2 . Чему равна координата тела через 15 секунд от начала движения?

Answer 1

$x = x_{0} + v_{0}t + \frac{at^2}{2}$

x₀ - начальная координата тела 
v₀ - скорость ( 3 м/c )
a - ускорение ( 0,2 м/c² )
t - время ( 15 с )

x₀ = v₀ * t₁ = 3 * 5 = 15 м 

$x = 15 + (3 * 15) + \frac{0,2 * 15^2}{2}$ = $15 + 45 + \frac{0,2 * 225}{2}$ = $60 + \frac{45}{2}$ = 60 + 22,5 = 82,5 м 

Answer 2

Дано Vo=3 м/с   t1=5  a=0,2 м/с2     t2=15-5=10 с    X- ?

X=X1+X2
X1=V0*t1=3*5=15 м

X2=V0*t2 +a*t2^2/2=3*10 +0,1*100=30+10=40 м

X=15+40=55 м

Ответ X=55 м

Answer 3

Calculation of the Body's Position

To calculate the position of the body after 15 seconds from the start of its motion, we need to consider two phases: the initial phase of constant velocity and the subsequent phase of acceleration.

1. Initial Phase (0 to 5 seconds): - The body moves with a constant velocity of 3 m/s for 5 seconds. - The formula to calculate the distance traveled during this phase is distance = velocity × time. - Therefore, the distance traveled during this phase is 3 m/s × 5 s = 15 m.

2. Subsequent Phase (5 to 15 seconds): - After 5 seconds, the body receives an acceleration of 0.2 m/s^2. - The formula to calculate the distance traveled during this phase is distance = initial velocity × time + 0.5 × acceleration × time^2. - The initial velocity during this phase is the final velocity of the previous phase, which is 3 m/s. - Therefore, the distance traveled during this phase is (3 m/s × 10 s) + (0.5 × 0.2 m/s^2 × (10 s)^2) = 30 m + 10 m = 40 m.

To find the total distance traveled by the body after 15 seconds, we sum up the distances from both phases: Total distance = distance during initial phase + distance during subsequent phase = 15 m + 40 m = 55 m.

Therefore, the coordinate of the body after 15 seconds from the start of its motion is 55 meters.

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