A 67 kg kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m m . What is the effective s
1 answer:
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Answer:
4.8 m/s
Explanation:
When she catches the train,
- They will have travelled the same distance.and
- Their speeds will be equal
The formula for the distance covered by the train is
d = ½at² = ½ × 0.40t² = 0.20t²
The passenger starts running at a constant speed 6 s later, so her formula is
d = v(t - 6.0)
The passenger and the train will have covered the same distance when she has caught it, so
(1) 0.20t² = v(t - 6.0)
The speed of the train is
v = at = 0.40t
The speed of the passenger is v.
(2) 0.40t = v
Substitute (2) into (1)
0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t
Subtract 0.20t² from each side
0.20t² - 2.4t = 0
Factor the quadratic
t(0.20t - 2.4) = 0
Apply the zero-product rule
t =0 0.20t - 2.4 = 0
0.20t = 2.4
(3) t = 12
We reject t = 0 s.
Substitute (3) into (2)
0.40 × 12 = v
v = 4.8 m/s
The slowest constant speed at which she can run and catch the train is 4.8 m/s.
A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.
Answer:
a) F = -1035.385 N
b) Backwards
c) s = 15.60 m
Explanation:
Given information
= Initial Speed of Car = 15.0 m/s
= Final Speed of Car = 9.00 m/s
= Breaking Time = 1.30 s
= Mass of Car = 1040 kg
Part (a)
To find the force exerted on the car we use the following formula
Where
= Force = unknown
= Mass of Car = 1040 kg
= Acceleration of Car / Deceleration of Car = unknown
To find the force (F) we need to first find the deceleration rate (a)
To find the deceleration rate we use the following formula
Inputting the given values
To find the force
Part (b)
Since the value of F is negative this means the the force was opposite the direction of motion, hence the force was backwards.
Part (c)
To find the total distance the car moved while braking we use the following formula
Where
= distance traveled
Inputting the values given