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a) Directly read the y-axis value of the graph when t = 0s.
b) Straight line graph for the time interval involved. This means that the speed (gradient) is constant. The modulus sign is introduced to remove the negative sign which would have been obtained.
c) Because the velocity is constant for the first 10 seconds, it means that acceleration (rate of change of velocity) is 0.
d) Might be helpful to draw a velocity-time graph. Since acceleration is constant, the graph would be a straight-line graph. Given the time interval asked in the question is for the last 10 seconds, the graph would have the shape as shown. DIstance can be read off the graph when t=15s. Given that the area under the velocity-time graph represents distance travelled, the speed when t=25s can be obtained by equating the area under the graph to 175m and solving for v.
b) Straight line graph for the time interval involved. This means that the speed (gradient) is constant. The modulus sign is introduced to remove the negative sign which would have been obtained.
c) Because the velocity is constant for the first 10 seconds, it means that acceleration (rate of change of velocity) is 0.
d) Might be helpful to draw a velocity-time graph. Since acceleration is constant, the graph would be a straight-line graph. Given the time interval asked in the question is for the last 10 seconds, the graph would have the shape as shown. DIstance can be read off the graph when t=15s. Given that the area under the velocity-time graph represents distance travelled, the speed when t=25s can be obtained by equating the area under the graph to 175m and solving for v.
Date Posted:
1 year ago
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