An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first
1 answer:
Answer:
The right answer is "1.369 m/s²".
Explanation:
The given values are:
Distance (s)
= 260 m
Initial speed (u)
= 26 m/s
Reaction time (t')
= 0.51 s
During reaction time, the distance travelled by locomotive will be:
⇒
Remained distance between locomotive and car:
⇒
Now,
The final velocity to avoid collection is, V = 0 m/s
From third equation of motion:
⇒
On putting the estimated values, we get
⇒
⇒
⇒
⇒
⇒
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Answer:
(i) Please find attached the required velocity time graphs plotted with MS Excel
(ii) The velocity of vehicle A at the 18th second = 20 m/s
The velocity of vehicle B at the 18th second = 0 m/s
(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters
Explanation:
The given parameters of the motion of vehicles A and B are;
The acceleration of vehicles A and B = Uniform acceleration starting from rest
The maximum velocity attained by vehicle A = 30 m/s
The time it takes vehicle A to attain maximum velocity = 10 s
The maximum velocity attained by vehicle B = 30 m/s
The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s
The time duration vehicle A maintains its maximum velocity = 6 s
The time duration vehicle B maintains its maximum velocity = 4 s
(i) From the question, we get the following table;
From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here
(ii) The velocity of vehicle A at the start = 0 m/s
After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s
The maximum velocity is maintained for 6 seconds which gives;
At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s
The time it takes vehicle A to decelerate to rest = 6 s
The deceleration of vehicle A, = (30 m/s - 0 m/s)/(6 s) = 5 m/s²
Therefore, we get;
v = u - ·t
At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s
u = 30 m/s
∴ v₁₈ = 30 - 5 × 2 = 20
The velocity of vehicle A at the 18th second, = 20 m/s
For vehicle B, we have;
At the 14th second, the velocity of vehicle B = 40 m/s
Vehicle B decelerates to rest in, t = 4 s
The deceleration of vehicle B, = (40 m/s - 0 m/s)/(4 s) = 10 m/s²
For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s
∴ = 40 m/s - 10 m/s² × 4 s = 0 m/s
The velocity of the vehicle B at 18th second, = 0 m/s
(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;
The area triangle A₁ = (1/2) × 10 × 30 = 150
Area of rectangle, A₂ = 6 × 30 = 180
Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50
The distance covered in the 18th second by vehicle = A₁ + A₂ + A₃
∴ = 150 + 180 + 50 = 380
The distance covered in the 18th second by vehicle = 380 m
The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;
Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440
The distance covered by the trapezoid, = 440 m
The distance of the two vehicles apart at the 18t second, =
-
∴ = 440 m - 380 m = 60 m
The distance of the two vehicles from one another at the 18th second, = 60 m.
Explanation:
The given data is as follows.
Mass of small bucket (m) = 4 kg
Mass of big bucket (M) = 12 kg
Initial velocity () = 0 m/s
Final velocity () = ?
Height = 2 m
and, = 0 m
Now, according to the law of conservation of energy
starting conditions = final conditions
235.44 = + 78.48
= 4.43 m/s
Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.