Physics, asked by Xiphodon, 10 months ago

A car A is travelling on a straight level road with a speed of 60km/hr. It is followed by another car B which
is moving with a speed of 70km/hr. when the distance between them is 2.5km, the car B is given a
deceleration of 20km/hr^2. After what distance will the car B catch up with car A??​

Answers

Answered by vaishanavi2003
17

Answer :

Distance covered by car A in time t is, s1 = 60 t

For car B:

Using the second equation of motion: s = ut + at2

where: s = distance covered = s2

u = initial velocity = 70 km/h

a = acceleration = -20 km/h

t = time = t

s = ut + at2

s2 = 70t + (-20) × t2

s2 = 70t – 10 t2

But it is given that s2 – s1 =2.5 km

70t – 10 t2 - 60 t = 2.5

10t - 10 t2 = 2.5

t – t2 = 0.25

t – t2 – 0.25 =0

t2 –t +0.25 =0

(t – 0.5)2 =0

t = 0.5 hr

Answered by VishalSharma01
94

Answer:

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 have

Speed of car A = 60 km/h

Speed of car B = 70 km/h

Distance between car B and A = 2.5 km

i.e D(B) - D(A) = 2.5 km

Acceleration (deceleration), a = - 20 km/h

To Find,

Time in which Car B will catch car A.

We know that,

Distance = speed × time

D(A) = 60 × T

D(A) = 60t km

And for Car B,

We know that,

s = ut + 1/2 × at²

⇒ s = 70 × t + 1/2 × (- 20) × t²

⇒ s = 70 - 10t² km

Then, D(B) =  70 - 10t² km

Its given that,

D(B) - D(A) = 2.5 km

⇒ 70t - 10t² - 60t = 2.5 km

⇒ 10t - 10t² = 2.5 km

By dividing eq by 10, we get

⇒ t - t² = 0.25

⇒ t - t² - 0.25 = 0

By 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.

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