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
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
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.