A cyclist starts from rest and coasts down a 4.0∘ hill. The mass of the cyclist plus bicycle is 85 kg. Ignore air resistance and
1 answer:
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Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course : = 0.75 m/s²,
= 20 m,
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s,
= -1.15 m/s²,
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) × )
0 = 129.96 - 2.3
2.3 = 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √( ² +
² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
Answer:
t = 13.7 s or t = 14 s with proper significant figures
Explanation:
The initial speed is 0 m/s since the car starts from rest, acceleration is 5.5 m/s2 and distance is 523 m.
Since we have initial speed, acceleration and distance we can use the following formula to find the time. We can now use algebra to work out our answer.
d = vt + at²
523 = (0)t + ()(5.5)t²
523 = 2.8t²
186.8 = t²
13.7 s = t
(t = 14 s with proper significant figures)