Answer: 10.63
Step-by-step explanation: 8.1 + 2.53
Step-by-step explanation:
It is given that,
Initial speed of the car, u = 40 m/s
Final speed of the car, v = 25 m/s
Time taken, t = 3.5 s
(a) We need to find the magnitude and direction of the car’s acceleration as it slows down. It can be calculated using formula as :
The acceleration is in the opposite direction of motion i.e. west.
(b) Let s is the distance the car travel in the 3.5-s time period. It can be calculated using the third equation of motion as


s = 113.9 meters
Hence, this is the required solution.
Answer:
t = -5
Step-by-step explanation:
Solve for t:
5 (t - 3) - 2 t = -30
Hint: | Distribute 5 over t - 3.
5 (t - 3) = 5 t - 15:
5 t - 15 - 2 t = -30
Hint: | Group like terms in 5 t - 2 t - 15.
Grouping like terms, 5 t - 2 t - 15 = (5 t - 2 t) - 15:
(5 t - 2 t) - 15 = -30
Hint: | Combine like terms in 5 t - 2 t.
5 t - 2 t = 3 t:
3 t - 15 = -30
Hint: | Isolate terms with t to the left hand side.
Add 15 to both sides:
3 t + (15 - 15) = 15 - 30
Hint: | Look for the difference of two identical terms.
15 - 15 = 0:
3 t = 15 - 30
Hint: | Evaluate 15 - 30.
15 - 30 = -15:
3 t = -15
Hint: | Divide both sides by a constant to simplify the equation.
Divide both sides of 3 t = -15 by 3:
(3 t)/3 = (-15)/3
Hint: | Any nonzero number divided by itself is one.
3/3 = 1:
t = (-15)/3
Hint: | Reduce (-15)/3 to lowest terms. Start by finding the GCD of -15 and 3.
The gcd of -15 and 3 is 3, so (-15)/3 = (3 (-5))/(3×1) = 3/3×-5 = -5:
Answer: t = -5
Answer:
Step-by-step explanation:
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The correct answer is d15