Question

A cyclist rides off from rest, accelerating at a constant rate for 3 minutes until she reaches 40 kmh-1. She then maintains a constant speed for 4 minutes until reaching a hill. She slows down at a constant rate over one minute to 30 kmh-1. then continues at this rate for 10 minutes.

At the top of the hill she reduces her speed uniformly and is stationary 2 minutes later.

b
How far has the cyclist travelled? Its 9.75 km, but I don't understand how to get there


PLEASE SHOW YOUR WORK

Answer 1

Explanation:

To solve this problem, we need to use the equations of motion for constant acceleration, constant velocity, and constant deceleration. We'll break the problem into several parts and use these equations to find the distance traveled in each part. Then, we'll add up the distances to get the total distance traveled.

First, we need to convert the units of speed from km/h to m/s, since the equations of motion use meters per second. We have:

Initial speed (u) = 0 km/h = 0 m/s

Final speed (v) = 40 km/h = 11.11 m/s

Constant speed = 40 km/h = 11.11 m/s (for 4 minutes)

Final speed before hill = 30 km/h = 8.33 m/s

Speed at top of hill = 0 m/s

Acceleration (a) = (v-u)/t = (11.11-0)/(3*60) = 0.0611 m/s^2

PART 1: ACCELERATION PHASE

Time taken (t) = 3 minutes = 180 seconds

Distance traveled (s) = ut + (1/2)at^2

s = 0 + (1/2)0.0611(180^2) = 331.83 meters

PART 2: CONSTANT SPEED PHASE

Time taken (t) = 4 minutes = 240 seconds

Distance traveled (s) = vt

s = 11.11*240 = 2666.4 meters

PART 3: DECELERATION PHASE

Time taken (t) = 1 minute = 60 seconds

Deceleration (a) = (v-u)/t = (8.33-11.11)/60 = -0.0461 m/s^2 (negative since it's deceleration)

Distance traveled (s) = vt + (1/2)at^2

s = 8.3360 + (1/2)(-0.0461)*(60^2) = 494.7 meters

PART 4: CONSTANT SPEED PHASE

Time taken (t) = 10 minutes = 600 seconds

Distance traveled (s) = vt

s = 8.33*600 = 4998 meters

PART 5: DECELERATION PHASE TO STOP

Time taken (t) = 2 minutes = 120 seconds

Initial speed (u) = 8.33 m/s

Final speed (v) = 0 m/s

Deceleration (a) = (v-u)/t = (0-8.33)/120 = -0.0694 m/s^2

Distance traveled (s) = vt + (1/2)at^2

s = 8.33120 + (1/2)(-0.0694)*(120^2) = 733.3 meters

TOTAL DISTANCE TRAVELED:

Adding up the distances from each part, we get:

Total distance = 331.83 + 2666.4 + 494.7 + 4998 + 733.3 = 9184.23 meters = 9.18 km (rounded to two decimal places)

Therefore, the cyclist has traveled approximately 9.18 km.