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

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point

a

step-by-step explanation:

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