Question

A cyclist accelerates uniformly from rest with an acceleration of 1.4 m s^-2 for 20seconds.The cyclist then travels at this constant velocity for 30 seconds before deceleratinguniformly to rest.The total distance travelled is 1260 metres.a) Sketch a velocity-time graph to show this information.Use the velocity-time graph to find each of the following.b)The greatest velocity of the cyclist.C)The total time taken for the journey.d) The acceleration of the cyclist during the finale)Find the average speed of the cyclist during the whole journey.

Answer 1

Given:

The cyclist starts from rest.

The acceleration is

`a1=\text{ 1.4 m/s}^2`

during the time interval, t1 = 20 s.

The cyclist then travels with this constant velocity for t2 = 30 s.

Lastly, decelerates to rest.

Also, the total distance is d = 1260 m.

To find

(a) Sketch the velocity-time graph

(b) Greatest velocity of the cyclist

(c) Total time taken for the journey.

(d) The acceleration of cyclist during final stage.

(e) Average speed of the cyclist during the whole journey.

Step-by-step explanation:

(a) In order to sketch the velocity-time graph, we need to find the velocity for each time duration.

For time t1, the velocity can be calculated as

`\begin{gathered} a1\text{ = }(v1-0)/(t1) \\ v1=\text{ a1}^* t1 \\ =1.4*20 \\ =28\text{ m/s} \end{gathered}`

For time t2, the velocity is constant whose value is 28 m/s.

The above diagram is the velocity-time graph.

Here, the vertical axis represents the velocity.

The horizontal axis represents the time.

During the time t1, velocity increases.

During time t2, velocity is constant.

During time t3, velocity decreases.

(b) From the graph, the highest value of velocity is 28 m/s.

Thus, the greatest velocity is 28 m/s.

(c) In order to calculate total time, we need to calculate the time during deceleration.

We have been provided with the value of total distance.

The total distance in a velocity-time graph is the area under the curve.

In the above graph, there are three cases:

Acceleration: The area of the triangle is

`\begin{gathered} A1\text{ = }(1)/(2)*\text{ t1}* v1 \\ =(1)/(2)*20*28 \\ =280\text{ m} \end{gathered}`

Constant velocity: The area of the rectangle is

`\begin{gathered} A2\text{ = t2}* v1 \\ =30*28 \\ =840\text{ m} \end{gathered}`

Deceleration: The area of the triangle will be

`\begin{gathered} A3\text{ = Total distance - \lparen A1+A2\rparen} \\ =1260-(280+840) \\ =\text{ 140 m} \end{gathered}`

Thus, the time t3 will be

`\begin{gathered} A3\text{ =}(1)/(2)* t3* v1 \\ t3=(2* A3)/(v1) \\ =(2*140)/(28) \\ =10\text{ s} \end{gathered}`

Hence the total time is

`\begin{gathered} t=t1+t2+t3 \\ =20+30+10 \\ =\text{ 60 s} \end{gathered}`

(d) The acceleration during the final stage is

`\begin{gathered} a3\text{ = }(0-v1)/(t3) \\ a3=-(28)/(10) \\ =-2.8\text{ m/s}^2 \end{gathered}`

(e) The average speed can be calculated as

`\begin{gathered} v_(av)=\frac{total\text{ distance}}{total\text{ time}} \\ =\text{ }(1260)/(60) \\ =\text{ 21 m/s} \end{gathered}`