If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be A. equal to the weigh
1 answer:
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Answer:
a.) a = 0 ms⁻²
b.) a = 9.58 ms⁻²
c.) a = 7.67 ms⁻²
Explanation:
a.)
Acceleration (a) is defined as the time rate of change of velocity
Given data
Final velocity = v₂ = 0 m/s
Initial velocity = v ₁ = 0 m/s
As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,
a = 0 ms⁻²
b.)
Given data
As the space shuttle start from rest, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 8 min = 480 s
By the definition of Acceleration (a)
a = 9.58 ms⁻²
c.)
Given data
As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 10 min = 600 s
By the definition of Acceleration (a)
a = 7.67 ms⁻²
Answer:
Explanation:
As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from to
is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.
The total distance covered by the train during the entire journey is the area of the speed-time graph.
Area
As velocity is in and time is in
so the unit of area is
Hence, the total distance is .
