A boat covered 5/9 of a journey at an average speed of 25
1 answer:
You might be interested in
Answer:
(A). The work done by friction in crossing the patch is -637.98 J.
(B). The speed of skier is 10.57 m/s.
Explanation:
Given that,
Mass of skier = 62 kg
Speed = 6.5 m/s
Length = 3.50 m
Coefficient kinetic friction = 0.30
Height = 2.5 m
(A) we need to calculate the work done by friction in crossing the patch
Using formula of work done
Put the value into the formula
The work done by friction in crossing the patch is -637.98 J.
(B) we need to calculate the speed of skier
Using conservation of energy
Final potential energy is zero
So,
Put the value into the formula
The speed of skier is 10.57 m/s.
Hence, (A).The work done by friction in crossing the patch is -637.98 J.
(B).The speed of skier is 10.57 m/s.
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.
