6 A 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 is moving with a speed of 70 km/h. When the distance between them is 2.5 km, the car B is given a deceleration of 20 km/h. After how much time will catch up with A?
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Speed of B = 70km/h
Initial velocity = 70km/h
Acceleration = -20km/h
Let time be t
Therefore, distance = ut + 1/2at²
= 70 × t + 1/2 × -20 × t²
= 70t - 10t²
Uniform speed of A = 60km/h
Therefore, distance = speed × time
= 60 × t
= 60t
Therefore, ATQ.
(70t - 10t²) - 60t = 2.5
=⟩ 10t - 10t² = 2.5 (Dividing every term by 10)
=⟩ t - t² = 25 /100
= 1 / 4
=⟩ t - t² - 1/4 = 0
=⟩ t² - t + 1/4 = 0
=⟩ (t - 1/2)² = 0
=⟩ t - 1/2 = √0
= 0
=⟩ t = 1/2hrs or 30mins
After 30mins B will will catch up with A.
Hope this helped you, cheers :)
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