Two cars A and B cross a point simultaneously with velocities 10 and 15 m/s . They move with different uniform accelerations. A overtakes B with a speed of 25 m/s. What is the velocity of B at that instant ?
Answers
Given that, two cars A and B cross a point simultaneously with velocity 10 m/s and 15 m/s.
Initial Velocity of car A = 10 m/s and Initial Velocity of car B = 15 m/s
Both the cars move with different uniform accelerations. A overtakes B with a speed of 25 m/s.(Final velocity of car A is 25 m/s.)
We have to find the velocity of B at that instant.
For car A:
v = u + at
25 = 10 + at
15 = at ...............(1st equation)
For car B:
V = U + At
V = 15 + At ............(2nd equation)
Also,
Distance travelled by car A = Distance travelled by car B (As both car are travelling a same distance or distance travelled by both cars is same)
ut + 1/2 at² = Ut + 1/2 At²
(10)t + 1/2 at² = (15)t + 1/2 At²
10 + 1/2 at = 15 + 1/2 At
10 - 15 = 1/2 (At - at)
-5*2 = At - at
-10 = At - at
At = at - 10
From 1st equation
At = 15 - 10 = 5 m/s
Substitute value of At in (2nd equation)
V = 15 + 5 = 20 m/s
Therefore, the velocity of the car B at the instant is 20 m/s.
Answer:
Explanation:
Given :-
Initial velocity of car A, u₁ = 10 m/s
Initial velocity of car B, u₂ = 15 m/s
Final velocity of car A, v₁ = 25 m/s
To Find :-
Final velocity of car B, v₁ = ??
Formula to be used :-
1st and 2nd equation of motion,
v = u + at and s = ut + 1/2 at²
Solution :-
Putting all the values, we get
v = u + at
⇒ v₁ = u₁ + a₁ × t
⇒ a₁ × t = v₁ - u₁
⇒ a₁ × t = 25 - 10
⇒ a₁ × t = 15 ..... (i)
Now, v = u + at
⇒ a₂ × t = u₂ - v₂
⇒ a₂ t + 15 = v₂ .... (ii)
Now, s = ut + 1/2 at²
Comparing both sides, we get
⇒ s = ut + 1/2 at² = s = ut + 1/2 at²
⇒ u₁ × t + 1/2 × a₁ × t² = u² × t + 1/2 × a₂ × t²
⇒ u₁ + 1/2 × a₁ × t = u₂ + 1/2 × a₂ × t
⇒ 10 - 15 = 1/2 (At - at)
⇒ - 5 × 2 = a₂t - a₁ t
⇒ - 10 = a₂t - a₁t
⇒ a₂t = a₁t - 10
⇒ a₂t = 15 - 10 [From Eq (i)]
⇒ a₂t = 5 m/s
⇒ 5 + 15 = v₂ [ From (ii)]
⇒ v² = 20 m/s
Hence, the velocity of B at that instant is 20 m/s.