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Confused about velocity-time graph QUESTION

Hi
unsure how to calculate one square of this graph for a curve.
help needed
Mark scheme
Evidence of determining the area under the graph 1 square is equivalent to 2.5 × 10−4 (m) OR 9 to 11 squares 2.3 × 10−3 to 2.7 × 10−3 (m)

a.JPG
2 squares on vertical axis represent 0.1 ms-1
so 1 square on vertical axis represents 0.05 ms-1

2 squares on horizontal axis represent 1 x 10^-2 s
1 square on horizontal axis represents 0.5 x 10^-2 s

so area of 1 square represents = 0.5 x 10^-2 x 0.05 = 2.5 x 10^-4 m

the required area is 2.5 x 10^-3 m (10 squares)
Reply 2
Avatar for Batman2k1
Batman2k1
OP
14
Thank you! A lot more clearer now
Original post by BobbJo
2 squares on vertical axis represent 0.1 ms-1
so 1 square on vertical axis represents 0.05 ms-1

2 squares on horizontal axis represent 1 x 10^-2 s
1 square on horizontal axis represents 0.5 x 10^-2 s

so area of 1 square represents = 0.5 x 10^-2 x 0.05 = 2.5 x 10^-4 m

the required area is 2.5 x 10^-3 m (10 squares)

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