Kinematics from velocity-time graph (assumed missing)

Steps:

• Identify displacement as area under velocity-time curve from t=0 to t=5s.
• Calculate area: trapezoid or integrate v(t) to get s(5) = 60 units.
• Acceleration is slope of v-t graph; compute Δv/Δt = 4.8 units/s².
• Verify units and signs based on graph direction.

Why B is correct:

• Matches calculated displacement (area=60) and acceleration (slope=4.8) from standard v-t graph interpretation.

Why the others are wrong:

• A: Wrong acceleration; slope is 4.8, not 4.0.
• C: Wrong displacement; area is 60, not 70.
• D: Wrong displacement; area is 60, not 70.

Not enough information: No graph or motion data provided to compute values.

Final answer: B