An object of mass m moving with initial speed of 15 m/s got an acceleration of 10 m/s^ after a constant force F is applied to it. How long does it take for the object to reach 90 meters from its initial position?
S=Ut+1/2at^2
90=15t+1/2(10)t^2
90=15t+5t^2
90/5=15t/5+5t^2/5
18=5t+t^2
(t+6)(t-3)
t=-6 t=3
So the time does not be negative so the answer is t=3
With acceleration of 10 m/s^2,and initial velocity of 15 m/s, the displacement from initial position is
X = 15 t + 5 t^2
If X = 90,
5t^2 + 15t -90 = 0
t^2 + 3t -18 = 0
(t+6)(t-3) = 0
The positive solution is t = 3 seconds. Ignore the negative solution; it is outside the domain of the function.
To determine how long it takes for the object to reach 90 meters from its initial position, we can use the kinematic equation:
\(s = ut + \frac{1}{2}at^2\)
where:
s = displacement (90 m)
u = initial speed (15 m/s)
a = acceleration (10 m/s^2)
t = time (unknown)
Rearranging the equation, we have:
\(90 = 15t + \frac{1}{2}(10)t^2\)
Simplifying further:
\(90 = 15t + 5t^2\)
Converting this equation to a quadratic equation in standard form:
\(5t^2 + 15t - 90 = 0\)
To solve the quadratic equation, we can factor it:
\(5(t^2 + 3t - 18) = 0\)
\(5(t + 6)(t - 3) = 0\)
Setting each factor equal to zero:
t + 6 = 0 or t - 3 = 0
Solving for t:
t = -6 or t = 3
Since time cannot be negative, the only valid solution is t = 3.
Therefore, it will take 3 seconds for the object to reach 90 meters from its initial position.
To determine the time it takes for the object to reach a distance of 90 meters from its initial position, we can use the equations of motion.
First, let's find the final velocity of the object using the equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 15 m/s
a = 10 m/s²
Substituting the known values into the equation, we can find the final velocity:
v = 15 m/s + 10 m/s² * t
Next, we need to find the time it takes for the object to reach a distance of 90 meters. We can use the equation:
s = ut + (1/2) * a * t²
where:
s = distance
u = initial velocity
a = acceleration
t = time
Given:
s = 90 m
u = 15 m/s
a = 10 m/s²
Substituting the known values into the equation, we get:
90 = 15 * t + (1/2) * 10 * t²
This equation is quadratic in form. Simplifying it, we have:
90 = 15t + 5t²
Rearranging the equation, we get:
5t² + 15t - 90 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring is not straightforward in this case, so let's use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
For our equation, a = 5, b = 15, and c = -90. Substituting these values into the formula, we get:
t = (-15 ± √(15² - 4*5*(-90))) / (2*5)
Simplifying further:
t = (-15 ± √(225 + 1800)) / 10
t = (-15 ± √2025) / 10
t = (-15 ± 45) / 10
This gives us two possible values for time:
t₁ = (-15 + 45) / 10 = 3 seconds
t₂ = (-15 - 45) / 10 = -6 seconds
Since time cannot be negative in this context, we discard the negative value.
Therefore, it will take the object 3 seconds to reach a distance of 90 meters from its initial position.