The figure shows the v-t graph of a truck moving on the road. The distance travelled in the first 8 seconds is

Question

BlackTom AI · Accepted Answer

The question asks for the distance travelled in the first 8 seconds, which is the area under the velocity-time (v-t) graph from t = 0 to t = 8 s.
Option 1 (15 m): This would correspond to a much smaller area, since the velocity is zero at t = 0 and increases linearly, so the area up to 8 s cannot be as low as 15 m.
Option 2 (30 m): Though closer, 30 m is still too small given the velocity reaches 12 m/s by t = 8 s (see below). The area under a linearly increasing v-t graph grows more than linearly with time; 30 m would require a much lower average speed than what the graph implies.
Option 3 (120 m): This would require an average velocity of 15 m/s over 8 s, which is impossible here because the velocity at t = 8 s is only 12 m/s (and the velocity starts from 0). Therefore 120 m is too large.
Option 4 (60 m): This is closer to the correct value but still not exact; we will see why it’s not correct when we compute precisely.

To compute accurately, note the graph shows a straight line from (0, 0) to (10 s, 15 m/s). The velocity at time t is v(t) = (15/10) t = 1.5 t for 0 ≤ t ≤ 10. The distance traveled from 0 to 8 s is the area under this line from 0 to 8:
- Using the triangle area method: the velocity at t = 8 s is v(8) = 1.5 × 8 = 12 m/s. The average velocity over 0–8 s is (0 + 12)/2 = 6 m/s. Distance = average velocity × time = 6 × 8 = 48 m.
- Alternatively, integrate: ∫(0 to 8) 1.5 t dt = 1.5 × (t^2/2) |0^8 = 0.75 × 64 = 48 m.
Thus, the exact distance traveled in the first 8 seconds is 48 m. The provided options do not include 48 m. The statement in the given answer field acknowledges that none of the options match the calculated distance, noting that 60 m is the closest but not accurate.

The figure shows the v-t graph of a truck moving on the road. The distance travelled in the first 8 seconds is Single choice

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