The 2020 Ford police interceptor has a max acceleration of 4.4 m/s2. If a policeperson is sitting on the roadside when a speeding car passes them at 70 mph (31.3 m/s), and the policeperson takes off
at max acceleration 1 second later, will they be able to catch the speeder in 10 seconds?
(Show work)

The policeperson will not be able to catch the speeder in 10 seconds

Explanation:

The maximum acceleration of the 2020 Ford police interceptor, a = 4.4 m/s²

The rate at which the speeding car is moving, v = 70 mph (31.3 m/s)

The time after which the policeperson takes off = 1 second

The time, t, at which the policeperson catches the speeding car is given as follows;

s = u·t + 1/2·a·t²

Where;

u = The initial velocity of the Ford police interceptor = 0

∴ s = 0 × t + 1/2·a·t² = 1/2·a·t²

s = 1/2·a·t²

s = v × (t + 1)

Equating both equations, gives;

v × (t + 1) = 1/2·a·t²

Substituting the values, we get;

31.3 × (t + 1) = 1/2 × 4.4 × t²

31.3·t + 31.3 = 2.2·t²

2.2·t² - 31.3·t - 31.3 = 0

By the quadratic formula, we have;

t = (31.3 ± √((-31.3)² - 4×2.2×(-31.3))/(2 × 2.2)

t ≈ -0.938 seconds or t ≈ 15.165 seconds

The policeperson will catch the speeder in approximately 15.165 seconds

Therefore, the policeperson will not be able to catch the speeder in 10 seconds.