Respuesta :
Answer:
After t = 100 second the car move to 3.50 km
Step-by-step explanation:
Given: A car that initially moving at 7.50 m/s begins to accelerate forward uniformly at 0.550 m/s.
here, initial velocity(v) = 7.50 m/s, acceleration(a)= 0.550 [tex]m/s^2[/tex] and [tex]\delta x =3.50 km[/tex] = 3500 m ( ∵At initial stage distance x = 0)
Using the kinematic formula for the change in distance [tex](\delta x)[/tex] of an object moving with initial velocity (v), and constant acceleration (a) i.e,
[tex]\delta x = vt+\frac{1}{2}a t^2[/tex]
Using the given information, we solve for t:
[tex]3500 = 7.50t +\frac{1}{2}\cdot (0.550)t^2[/tex] or
[tex]3500 = 7.50t +0.275t^2[/tex]
Multiply both sides by 1000 we get;
3500000= 7500t+275t^2 or
[tex]275t^2+7500t-3500000=0[/tex] or
[tex]11t^2+300t-140000=0[/tex] ......[1]
Solve with the quadratic formula:
For a quadratic equation of the form [tex]ax^2+bx+c=0[/tex]; the solutions are ;
[tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}} {2a}[/tex].
From the equation [1], we have
the value of a=11 , b= 300 and c=140000
Substitute the value in above quadratic formula:
then,
[tex]t_{1,2}=\frac{-300\pm\sqrt{(300)^2-4(11)(140000)}} {2(11)}[/tex]
On solving we get, the value of t:
t= 100 s (time will not be negative)
Therefore,
t = 100s long after beginning to accelerate does the car to move 3.50 km
