Answer:
a) The total displacement of the trip was 5.32 × 10³ m
b) The average speeds were:
leg 1: 24.5 m/s
leg 2: 49 m/s
leg 3: 23.9 m/s
Complete trip: 43.8 m/s
Explanation:
The position and velocity equations for an object moving along a straight line are as follows:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
If the velocity is constant, then a = 0 and x = x0 + v · t where "v" is the velocity.
a) To calculate the total displacement of the trip, let´s calculate the distance traveled in each phase.
Phase 1:
x = x0 + v0 · t + 1/2 · a · t²
x = 0 m + 0 m/s · t + 1/2 · 2.45 m/s² · (20.0 s)²
x = 490 m
The velocity reached in that phase is:
v = v0 + a · t
v = 0 m/s + 2.45 m/s² · 20.0 s
v = 49.0 m/s
Phase 2:
x = x0 + v · t
x = 490 m + 49.0 m/s · 96.0 s
x = 5.19 × 10³ m
Phase 3:
x = x0 + v0 · t + 1/2 · a · t²
x = 5.19 × 10³ m + 49 m/s · 5.44 s - 1/2 · 9.00 m/s² · (5.44 s)²
x = 5.32 × 10³ m
The total displacement of the trip was 5.32 × 10³ m
b) The average speed is calculated as the traveled distance divided by the elapsed time:
average speed v = final position - initial position / (final time- initial time)
Phase 1:
v = 490 m - 0 m / 20.0 s = 24.5 m/s
Phase 2:
v = 5.19 × 10³ m - 490 / 96.0 s
v = 48.9 m/s (without rounding the final position the result is 49.0 m/s)
Phase 3:
v = 5.32 × 10³ m - 5.19 × 10³ m / 5.44 s = 23.9 m/s
For the complete trip:
v = 5.32 × 10³ m - 0 m / (20.0 s + 96.0 s + 5.44 s)
v = 43.8 m/s
