The area-
The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.
<span><span>Area of light-blue triangle -
<span>The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: <span>area of triangle = 1⁄2 × base × height </span><span>so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. </span></span></span><span> Area of dark-blue rectangle
The width of the rectangle is 6 seconds and the height is 8 meters per second. So the area is 8 × 6 = 48m.</span><span> Area under the whole graph
<span>The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.<span>This is the total area under the distance-time graph. This area represents the distance covered.</span></span></span></span>
<h2>
Time taken is 0.459 seconds</h2>
Explanation:
We have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = 81 km/hr = 22.5 m/s
Time, t = ?
Acceleration, a = 49 m/s²
Substituting
v = u + at
22.5 = 0 + 49 x t
t = 0.459 seconds
Time taken is 0.459 seconds
Answer:
Action - Pulling up the train.
Reaction - Friction on the locomotive
Explanation:
Locomotive is pulling the train upwards ,
Which is the action force applied by the locomotive,
As a reaction locomotive will be pulled by the train which is the reaction of pulling
Now, considering it as a action on locomotive , friction force will act on it as a reaction upwards which will result to move it upwards.
For train action is pulling up by locomotive and reaction will be friction acting on it downwards.
The question looks incomplete, but according to the information given above seem like they have <span>identical journeys.
</span>a. the displacement of car A - <span> 65.5 m
</span>b. the displacement of car B - <span>65.5 m
c. average velocir</span>y of A
d. the average velocity of car B has the same.
