A car moving on a straight highway with a speed of V0 = 72 km / h, starting overtaking, accelerates with

A car moving on a straight highway with a speed of V0 = 72 km / h, starting overtaking, accelerates with constant acceleration. Find the vehicle speed module after the time t = 10 s of acceleration, if in the last two seconds of movement it covered the distance S = 58m. Also define the vehicle acceleration module A.

Speed ​​72 km / h = 72,000 m / 3600 s = 20 m / s.
Let’s denote:
speed at the beginning of overtaking – v₀ = 20 m / s,
speed after 8 and 10 seconds – V8 and V₁₀,
acceleration – a,
way in 8 and 10 seconds – S₈ and S₁₀.
We rewrite the general equation for the path s = v₀t + at² / 2 for the 8th and 10th seconds:
S₈ = 20 m / s * 8 s + a * 64 s² / 2 = 160 m + a * 32 s²
S₁₀ = 20 m / s * 10 + a * 100 s² / 2 = 200 m + a * 50 s²
Subtract the first equality from the second:
S₁₀ – S8 = 40 m – a * 18 s².
Substitute the condition S₁₀ – S8 = 58 m:
58 m = 40 m + a * 18 s²;
a = 1 m / s².
Velocity at any moment of time v = v₀ + at. Speed ​​after 10 s:
V₁₀ = 20 m / s * 10 s + 1 m / s² * 10 s = 20 m / s + 10 m / s = 30 m / s.

Answer: modulus of speed – 30 m / s, modulus of acceleration – 1 m / s².