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Answer:
a) v = 12.21m/s
a = 4.07 m/s²
b)v = 11.24m/s
a = 3.75 m/s²
Step-by-step explanation:
a) Dividing the moviment into two parts:
I - With acceleration
v = v₀ + at
s = s₀ + v₀t + at²/2
- v₀ = 0
- s₀ = 0
- a = ?
- v = ?
- t = 3s
- s = x
v = v₀ + at
v = 3a ⇒ a = v/3
s = s₀ + v₀t + at²/2
x = v/3.3²/2
x = 3v/2
II - Uniform
s = s₀ + vt
s = 100
s₀ = x
v = v
t = 9.69 - 3 = 6.69s
s = s₀ + vt
100 = x + v*6.69
100 = x + 6.69v
As x = 3v/2
100 = 3v/2 + 6.69v
100 = 1.5v + 6.69v
100 = 8.19v
v = 12.21m/s
a = v/3 = 4.07 m/s²
b) Dividing the moviment into two parts:
I - With acceleration
v = v₀ + at
s = s₀ + v₀t + at²/2
- v₀ = 0
- s₀ = 0
- a = ?
- v = ?
- t = 3s
- s = x
v = v₀ + at
v = 3a ⇒ a = v/3
s = s₀ + v₀t + at²/2
x = v/3.3²/2
x = 3v/2
II - Uniform
s = s₀ + vt
s = 200
s₀ = x
v = v
t = 19.30 - 3 = 16.30s
s = s₀ + vt
200 = x + v*16.3
100 = x + 16.3v
As x = 3v/2
200 = 3v/2 + 16.3v
200 = 1.5v + 16.3v
200 = 17.8v
v = 11.24m/s
a = v/3 = 3.75 m/s²
Answer:first add 14.6 to both side of the equation and that will give you what is x equal to
Step-by-step explanation:am just smart
Answer:
Step-by-step explanation:
Given
There are 52 cards in total
there are total of 13 pairs of same cards with each pair containing 4 cards
Probability of getting a pair or three of kind card=1-Probability of all three cards being different
Probability of selecting all three different cards can be find out by selecting a card from first 13 pairs and remaining 2 cards from remaining 12 pairs i.e.
for first card there are 52 options after choosing first card one pair is destroyed as we have to select different card .
For second card we have to select from remaining 12 pairs i.e. 48 cards and so on for third card.
Required Probability is 
