Question

Position, displacement, and distance A projectile is launched vertically from the ground at t=0, and its velocity in flight (in m/s) is given by v(t)=20−10t. Find the position, displacement, and distance traveled after t seconds, for 0≤t≤4.

Accepted Answer

Identify the given velocity function: $$v(t) = 20 - 10t$$, which represents the velocity of the projectile at time $$t$$ seconds. Find the position function $$s(t) by $$integrating the velocity function with respect to time: $$s(t) = \int v(t) \, dt = \int (20 - 10t) \, dt. $$Remember to include the constant of integration and use the initial condition $$s(0) = 0$$ since the projectile starts from the ground. Calculate the displacement after $$t$$ seconds by evaluating the position function at $$t=4$$: $$	ext{Displacement} = s(4) - s(0). $$Since $$s(0) = 0$$, this simplifies to $$s(4). $$Determine the distance traveled by considering the total length of the path. Since velocity changes sign, find the time when $$v(t) = 0 to $$identify when the projectile changes direction: solve $$20 - 10t = 0$$ for $$t. $$Then, calculate the position at this time and use it to find the total distance traveled by summing the absolute values of position changes over each interval. Summarize the results: position function $$s(t)$$ for $$0 \leq t \leq 4$$, displacement as $$s(4)$$, and distance traveled as the sum of absolute position changes considering the change in direction.

Position, displacement, and distance A projectile is launched vertically from the ground at t=0, and its velocity in flight (in m/s) is given by v(t)=20−10t. Find the position, displacement, and distance traveled after t seconds, for 0≤t≤4.

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Key Concepts

Velocity and Its Relation to Position

Displacement vs. Distance Traveled

Integration and Absolute Value in Calculating Distance