Using the provided graph and equation, answer the following questions about a cyclist's journey. What time…

Question

Using the provided graph and equation, answer the following questions about a cyclist's journey. What time did the cyclist leave home? When did they return? What was their distance from home at 10 a.m

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Here are the solutions to the questions based on the provided graph and equation. a) i. Step 1: Identify the starting point of the journey (distance from home is 0 km). The graph starts at (8 a.m., 0 km). The cyclist left home at 8 a.m. ii. Step 1: Identify the point where the cyclist returns home (distance from home is 0 km). The graph ends at (5 p.m., 0 km). The cyclist returned home at 5 p.m. iii. Step 1: Locate 10 a.m. on the time axis and find the corresponding distance. At 10 a.m., the distance from home was 20 km. iv. Step 1: Locate 11 a.m. on the time axis and find the corresponding distance. At 11 a.m., the distance from home was 20 km. v. Step 1: Locate 2 p.m. on the time axis and find the corresponding distance. At 2 p.m., the distance from home was 60 km. vi. Step 1: Locate 3 p.m. on the time axis and find the corresponding distance. At 3 p.m., the distance from home was 60 km. vii. Step 1: Identify horizontal segments on the graph, which indicate no change in distance over time. The cyclist took a rest from 10 a.m. to 11 a.m. and from 2 p.m. to 3 p.m. viii. Step 1: Locate noon (12 p.m.) on the time axis and find the corresponding distance. At noon, the distance from home was 40 km. ix. Step 1: Locate 5 p.m. on the time axis and find the corresponding distance. At 5 p.m., the distance from home was 0 km. x. Step 1: Calculate the speed (gradient) for each moving segment. Speed is the magnitude of the gradient of the distance-time graph. Segment 1 (8 a.m. to 10 a.m.): Speed₁ = 20 km − 0 km10 a.m. − 8 a.m. = 20 km2 h = 10 km/h Segment 2 (11 a.m. to 2 p.m.): Speed₂ = 60 km − 20 km2 p.m. − 11 a.m. = 40 km3 h ≈ 13.33 km/h Segment 3 (3 p.m. to 5 p.m.): Speed₃ = | 0 km − 60 km5 p.m. − 3 p.m. | = | −60 km2 h | = 30 km/h Step 2: Compare the speeds to find the fastest. The speeds are 10 km/h, 13.33 km/h, and 30 km/h. The fastest speed is 30 km/h. The cyclist was travelling most quickly from 3 p.m. to 5 p.m. b) Step 1: Rearrange the equation 6x − 2y − 4 = 0 into the slope-intercept form y = mx + c. 6x − 2y − 4 = 0; −2y = −6x + 4 Step 2: Divide by −2. y = (−6x)/(−2) + (4)/(−2); y = 3x − 2 Step 3: Identify the gradient (m) and y-intercept (c). The gradient is m = 3. The y-intercept is c = −2. The gradient is 3 and the y-intercept is −2.

Using the provided graph and equation, answer the following questions about a cyclist's journey. What time did the cyclist leave home? When did they return? What was their distance from home at 10 a.m

Answer

Step-by-step solution

Part a(i)

Part a(ii)

Part a(iii)

Part a(iv)

Part a(v)

Part a(vi)

Part a(vii)

Part a(viii)

Part a(ix)

Part a(x)· 1 step

Compare the speeds to find the fastest.

Part b· 3 steps3

Rearrange the equation $6x - 2y - 4 = 0$ into the slope-intercept form $y = mx + c$ .

Divide by $-2$ .

Identify the gradient ( $m$ ) and y-intercept ( $c$ ).

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More Mathematics Questions

Using the provided graph and equation, answer the following questions about a cyclist's journey. What time did the cyclist leave home? When did they return? What was their distance from home at 10 a.m

Answer

Step-by-step solution

Part a(i)

Part a(ii)

Part a(iii)

Part a(iv)

Part a(v)

Part a(vi)

Part a(vii)

Part a(viii)

Part a(ix)

Part a(x)· 1 step

Compare the speeds to find the fastest.

Part b· 3 steps3

Rearrange the equation 6x−2y−4=06x - 2y - 4 = 06x−2y−4=0 into the slope-intercept form y=mx+cy = mx + cy=mx+c.

Divide by −2-2−2.

Identify the gradient (mmm) and y-intercept (ccc).

Try one yourself

Still stuck on this one?

More Mathematics Questions

Rearrange the equation $6x - 2y - 4 = 0$ into the slope-intercept form $y = mx + c$ .

Divide by $-2$ .

Identify the gradient ( $m$ ) and y-intercept ( $c$ ).