Acar that initially moving at 7.50 m/s begins to accelerate forward uniformly at 0.550 m/s. how long after beginning to accelerate does it take the car to move 3.50 km

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 and = 3500 m    ( ∵At initial stage distance x = 0)

Using the kinematic formula for the change in distance of an object moving with initial velocity (v), and constant acceleration (a) i.e,

Using the given information, we solve for t:

or

Multiply both sides by 1000 we get;

3500000= 7500t+275t^2 or

 or

         ......[1]

Solve with the quadratic formula:

For a quadratic equation of the form ; the solutions are ;

.

From the equation [1], we have

the value of a=11 , b= 300 and c=140000

Substitute the value in above quadratic formula:

then,

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

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square

answer: asdfghjkl,mnbv

step-by-step explanation: