Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.
Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s
Answer: 11.4 s
Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²
Answer: 44.7 m/s²
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Given:-
- Time taken by the particle (t) = 6 s
- Average speed (v) = 40 m/s
To Find: Distance (s) travelled by the particle.
We know,
s = vt
where,
- s = Distance travelled,
- v = Speed &
- t = Time taken.
Putting the values,
s = (40 m/s)(6 s)
→ s = 240 m ...(Ans.)
Answer:
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