Question

Activity 2B-1 v (m/s) 1. A student rides a motorcycle along a straight road as described by the velocity vs. time graph shown to the right. At each of the following times, determine the motorcycle's position (relative to the motorcycle's initial location) a. 2 1 23 4 b. At each of the following times, determine the motorcycle's acceleration c. Two students are discussing their results from the part b) above: Student 1: Iget that the acceleration of the motorcycle at t -5.0 sec is zero since the velocity of the Student 2: But acceleration is about how the velocity changes, so to find acceleration we need to find the motorcycle is zero at that time. slope of the graph at t-5 s, and the slope is not zero there. Although the velocity is zero for that instant, it is changing from positive to zero to negative Which of these students, if any, do you agree with? Briefly explain your reasoning.

*please show all work and step by step solution

Answer 1

Let us divide the motion into 3 parts:

Part 1 : From time t = 0 to t =2 sec

Part 2 : From time t = 2 to t = 4 sec

Part 3 : From time t = 4 to t = 6 sec

t₁ = 0 sec

t₂ = 2 sec

t₃ = 4 sec

t₄ = 6 sec

Part 1:

Velocity increases from 0 to 6m/s from t=0 to t= 2sec

V₁ = 0 m/s at t = 0 sec

V₂ = 6 m/s at t = 2 sec

From the graph the velocity is a straight line during part 1 which implies that the motorcycle is under constant acceleration and the magnitude of the acceleration is given by the slope of the line of velocity.

Slope of line = (y₂ - y₁) / (x₂ - x₁)

Acceleration in part 1 = a₁

$a_{1} = \frac{V_{2} - V_{1}}{t_{2} - t_{1}}$

$a_{1} = \frac{6 - 0}{2 - 0}$

a₁ = 3 m/s²

Distance covered by the motorcycle in part 1 = d₁

V₂² = V₁² + 2a₁d₁

6² = 0² + 2x3(d₁)

d₁ = 6 m

The acceleration at any instant during part 1 that is from t=0 to t=2 sec is given by a₁ which is 3 m/s²

Part 2:

Velocity remains constant at 6m/s from t=2 to t=4 sec

V₂ = 6 m/s at t=2 sec

V₃ = 6 m/s at t=4 sec

As we observe during part 2 that the velocity is a straight line parallel to the x-axis that is the velocity is not changing it implies that the motorcycle is under zero acceleration.

Acceleration in part 2 = a₂ = 0 m/s²

Distance covered by the motorcycle in part 2 = d₂

d₂ = V₂(t₃ - t₂)

d₂ = 6(4-2)

d₂ = 12 m

The acceleration at any instant during part 2 that is from t=2 to t=4 sec is given by a₂ which is 0 m/s²

Part 3:

Velocity decreases from 6m/s to -6m/s from t=4 to t=6 sec

V₃ = 6 m/s at t = 4 sec

V₄ = -6 m/s at t = 6 sec

From the graph the velocity is a straight line during part 3 which implies that the motorcycle is under constant acceleration and the magnitude of the acceleration is given by the slope of the line of velocity.

Slope of line = (y₂ - y₁) / (x₂ - x₁)

Acceleration in part 3 = a₃

$a_{3} = \frac{V_{4} - V_{3}}{t_{4}- t_{3}}$

$a_{3} = \frac{-6 - 6}{6-4}$

a₃ = -6 m/s²

Distance covered by motorcycle in part 3 = d₃

V₄² = V₃² + 2a₃d₃

(-6)² = 6² + 2(-6)d₃

d₃ = 0 m

The acceleration at any instant during part 3 that is from t=4 to t=6 sec is given by a₃ which is -6 m/s²

a) Motorcycle's position relative to initial position :

At t = 2 sec = d₁ = 6m

At t = 4 sec = d₁ + d₂ = 6 + 12 = 18 m

At t = 6 sec = d₁ + d₂ + d₃ = 6 + 12 + 0 = 18 m

b) Motorcycle's acceleration

At t = 1 sec = 3 m/s²

At t = 3 sec = 0 m/s²

At t = 5 sec = -6 m/s²

c) Student 2 is correct in his explanation.

Acceleration is the rate of change of velocity with time and as per the graph given at t=5 sec the velocity is going from positive to zero to negative that is the velocity is changing and as the velocity is changing the acceleration cannot be zero.

Acceleration is zero only when the velocity is not changing.

The magnitude of acceleration can be given by the slope of the line as it depicts the rate of change of velocity with time.