The average speed of a particle in a time interval is defined as the distance travelled by the particle divided by the time interval. If particle travels a distance **s** in time **t _{1}** to

**t**, the average speed is defined as

_{2}The average speed gives the overall rapidity with which the particle moves in this interval. In a one-day cricket match, the average run rate is quoted as the total runs divided by the total number of overs used to make these runs.

Some of overs may be expensive and some may be economical meaning in some overs player may have scored higher runs while in others player may have scored lower runs.

So the average speed gives the total effect in the given interval. The rapidity or slowness total effect in the given interval.

Similarly when an athlete starts running, he or she runs slowly and gradually increases the rate. We define the instantaneous speed at a time t as follow.

Let **Δs** be the distance travelled in the time interval **t** to **t + Δt**

The average speed in this time interval is

Now make Δt vanishingly small and look for the value of Δs/Δt.

Remember Δs is the distance travelled in the chosen time interval Δt

As Δt approaches 0, the distance Δs also approaches zero but ration Δs/Δt also approaches zero but the ratio Δs/Δt has a **finite limit**.

Thus the **instantaneous speed** at a time t is defined as

where s is the distance travelled in time t.**The average speed is defined for a time interval and the instantaneous speed is defined at a particular instant.**

Let’s now understand difference between Average Speed and Instantaneous Speed using an example problem.

ProblemThe distance travelled by a particle in time t is given by s = (2.5 m/s^{2}) t^{2}Find 1) The average speed of particle during the time 0 to 5.0 s 2) The instantaneous speed at t = 5.0 s |

Solution1) The average speed of particle during the time 0 to 5.0 s As we discussed above Average Speed = Total Distance Covered divided by Total Time Taken It’s clear that Total Time Taken = 5.0 – 0 = 5.0 sBut we need to find out Total Distance Covered As per problem its given that s = 2.5 t^{2}At t = 0 s = 0 At t = 5.0 s = 2.5 × (5.0) ^{2} = 2.5 × 25.0 = 62.5s = 62.5 m Thus Total Distance Covered from t = 5.0 s to t = 0 is 62.5 mThus Total Distance covered = 62.5 m Total Time Taken = 5.0 s Average Speed = 62.5 / 5.0 = 12.5 m/s Thus average speed of particle during the time 0 to 5.0 s will be 12.5 m/s |

Solution2) The instantaneous speed at t = 5.0 s As we discussed above Instantaneous Speed = ds/dt As per problem we have s = 2.5 t^{2}Using rules of Differentiation ds/dt = 2.5 × 2 × t = 5.0t ds/dt = 5.0t So Instantaneous Speed = ds/dt = 5.0t So Instantaneous Speed at time t = 5.0 s will be 5.0 × 5.0 = 25.0 m/s Thus Instantaneous Speed at t = 5.0 s will be 25.0 m/s |

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