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mixer [17]
3 years ago
5

47. A car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. What

was the average speed of the car for the trip? a. 39 km/h b. 95 km/h c. 114 km/h d. 285 km/h
Physics
1 answer:
Otrada [13]3 years ago
3 0

Answer: 114 km/h

Explanation:

The formula for determining average speed is expressed as

Average speed = total distance/total time

The car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. It means that the total distance that the car travels is

85 + 200 = 285 km

The total time spent by the car is

0.5 + 2 = 2.5 hours

Therefore,

Average speed = 285/2.5 = 114 km/h

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Which term describes the number of crests that pass a point in a given amount of time
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Horizontal abduction involves the movement of the arm or thigh in the transverse plane from a(n) _____ position to a lateral pos
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Which of the following was NOT a main trade zone of the Indian Ocean Basin?
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6 0
3 years ago
A rocket accelerates upward from rest, due to the first stage, with a constant acceleration of a1 = 67 m/s2 for t1 = 39 s. The f
Igoryamba

Answer:

(a) v_1= a_1t_1

(b) v_2 =a_1t_1+a_2t_2

(c) 44133.5 m

Explanation:

<u>Given:</u>

  • u = initial speed of the rocket in the first stage = 0 m/s
  • v_1 = final speed of the rocket in the first stage
  • v_2 = final speed of the rocket in the second stage
  • t_1 = time interval of the first stage
  • t_2 = time interval of the second stage
  • s_1 = distance traveled by the rocket in the first stage
  • s_2 = distance traveled by the rocket in the second stage
  • s = distance traveled by the rocket in whole time interval

Part (a):

Since the rocket travels at constant acceleration.

\therefore v_1 = u+a_1t_1\\\Rightarrow v_1 = a_1t_1

Hence, the expression of the rocket's speed at time t_1\ is\ v_1 = a_1t_1.

Part (b):

In this part also, the rocket moves with a constant acceleration motion.

\therefore v_2 = v_1+a_2t_2\\\Rightarrow v_2 = a_1t_1+a_2t_2

Hence, the expression of the rocket's speed in the time interval t_2 is v_2 = a_1t_1+a_2t_2.

Part (c):

For the constant acceleration of rocket, let us first calculate the distance traveled by the rocket in both the time intervals.

s_1 = u+\dfrac{1}{2}a_1t_1^2\\\Rightarrow s_1 = 0+\dfrac{1}{2}67\times(1)^2\\\Rightarrow s_1 =33.5\ m

Similarly,

s_2 = v_1t_2+\dfrac{1}{2}a_2t_2^2\\\Rightarrow s_2 = a_1t_1t_2+\dfrac{1}{2}a_2t_2^2\\\Rightarrow s_2 = 67\times1\times49+\dfrac{1}{2}\times 34\times(49)^2\\\Rightarrow s_2 =44100\ m\\\therefore s = s_1+s_2\\\Rightarrow s = 33.5\ m+44100\ m\\\Rightarrow s =44133.5\ m

Hence, the rocket moves a total distance of 44133.5 m until the end of the second period of acceleration.

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3 years ago
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