Answer:
The time it will take the car to travel the next 15 m is 1 s.
Explanation:
Given;
initial velocity of the car, u = 0
time of motion, t = 2.4 s
initial distance traveled, d = 15 m
determine the constant acceleration of the car;
d = ut + ¹/₂at²
15 = 0 + (0.5 x 2.4²)a
15 = 2.88a
a = 15 / 2.88
a = 5.21 m/s²
The velocity of the car at the end of the first 15 m
v - u + at
v = 0 + 5.21 x 2.4
v = 12.5 m/s
The time to cover the next 15 m;
d₂ = vt + ¹/₂at²
15 = 12.5t + (0.5 x 5.21)t²
15 = 12.5t + 2.61t²
0 = 2.61t² + 12.5t - 15
Use the formula method to solve the above quadratic equation.
a = 2.61
b = 12.5
c = -15
[tex]t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-12.5 \ \ +/- \ \ \sqrt{12.5^2 - (4\times 2.61 \times -15)} }{2\times 2.61}\\\\t = 0.9937 \ s \ \ \ or \ \ \ -5.78 \ s\\\\time \ can \ only \ be \ positive;\\\\t \approx \ 1\ s[/tex]
Therefore, the time it will take the car to travel the next 15 m is 1 s.
