The correct graph is <u>D</u>.
The graph <em>A</em> is a straight line sloping downwards and it shows that the speed of the body is decreasing at a constant rate. Therefore, this s a graph of a body that is under a constant deceleration.
The graph B is a straight line which slopes upwards. Hence the graph shows that the speed of the body increases at a constant rate. Therefore, this is a graph of a body that is accelerating at a constant rate.
The graph C is curved line, which curves upwards. The slope of the curve increases with time. This is therefore, a graph of a body which is under increasing acceleration.
The graph D, however is a straight line parallel to the time axis. The speed of the body has the same value at all times. Therefore, Graph D is the graph which shows the motion of a body with constant speed.
Answer:
compressions; rarefactions
Explanation:
Answer:
1. The final velocity of the truck is 15 m/s
2. The distance travelled by the truck is 37.5 m
Explanation:
1. Determination of the final velocity
Initial velocity (u) = 0 m/s
Acceleration (a) = 3 m/s²
Time (t) = 5 s
Final velocity (v) =?
The final velocity of the truck can be obtained as follow:
v = u + at
v = 0 + (3 × 5)
v = 0 + 15
v = 15 m/s
Therefore, the final velocity of the truck is 15 m/s
2. Determination of the distance travelled
Initial velocity (u) = 0 m/s
Acceleration (a) = 3 m/s²
Time (t) = 5 s
Distance (s) =?
The distance travelled by the truck can be obtained as follow:
s = ut + ½at²
s = (0 × 5) + (½ × 3 × 5²)
s = 0 + (½ × 3 × 25)
s = 0 + 37.5
s = 37.5 m
Therefore, the distance travelled by the truck is 37.5 m
Answer:
v = 3.27 m/s
Explanation:
KE = 1/2 mv^2
695 J = 1/2 (130kg)(v^2)
695 J / (1/2 x 130kg) = v^2
v^2 = square root of 10.69
v = 3.27 m/s