Question

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.45 m/s^2 for 20.0 s. 2. Maintain a constant velocity for the next 1.60 min. 3. Apply a constant negative acceleration of −9.00 m/s^2 for 5.44 s. (a) What was the total displacement for the trip? _______m

(b) What were the average speeds for
legs 1____m/s
Leg 2____m/s
Leg 3_____m/s
complete trip_______m/s

Answer 1

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

Step-by-step 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