Explanation:Correct QUESTION,CAR A is travelling on a STRAIGHT level road with a UNIFORM speed of 60 km/h. It is followed by another car B which in moving with a speed of 70 km/h. When the distance between then is 2.5 km, the car B is given a deceleration of 20 km/h² . After how much time will B catch up with A.Solution,Here, we haveSpeed of car A = 60 km/hSpeed of car B = 70 km/hDistance between car B and A = 2.5 kmi.e D(B) - D(A) = 2.5 kmAcceleration (deceleration), a = - 20 km/hTo Find,Time in which Car B will catch car A.We know that,Distance = speed × timeD(A) = 60 × TD(A) = 60t kmAnd for Car B,We know that,s = ut + 1/2 × at²⇒ s = 70 × t + 1/2 × (- 20) × t²⇒ s = 70 - 10t² kmThen, D(B) =  70 - 10t² kmIts given that,D(B) - D(A) = 2.5 km⇒ 70t - 10t² - 60t = 2.5 km⇒ 10t - 10t² = 2.5 kmBy DIVIDING eq by 10, we get⇒ t - t² = 0.25⇒ t - t² - 0.25 = 0By factorization method, we get⇒ t² - t + 0.25 = 0⇒ t² - 0.5t - 0.5t + 0.25 = 0⇒ t(t - 0.5) - 0.5(t - 0.5) =0⇒ (t - 0.5) (t - 0.5) = 0⇒ (t - 0.5)² = 0⇒ t = 0.5 = 1/2⇒ t = 1/2 hours.Hence, Car B will catch car A is 1/2 hours.