Question

Starting from rest, a car takes 2.4 s to travel the first 15 m. Assuming a constant acceleration, how long will it take the car to travel the next 15 m?

Answer 1

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

$t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-12.5 \ \ +/- \ \ \sqrt{12.5^2 - (4\times 2.61 \times -15)} }{2\times 2.61}\\\\t = 0.9937 \ s \ \ \ or \ \ \ -5.78 \ s\\\\time \ can \ only \ be \ positive;\\\\t \approx \ 1\ s$

Therefore, the time it will take the car to travel the next 15 m is 1 s.