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`        An athelete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of his average acceleration?`
6 years ago

```							Sol. Initial velocity u = 0 (∴ starts from rest)
Final velocity v = 18 km/hr = 5 sec
(i.e. max velocity)
Time interval t = 2 sec.
∴ Acceleration = aave = (v-u )/t = 5/2 = 2.5 m/s2.

```
6 years ago
```							This question is very simple. You just need to find average acceleration. Formula:v-u/tU=0V=18×5/18=5 m/sT=2sa=5/2 =2.5 m/s2
```
3 months ago
```							Initial velocity=u= 0 m/sFinal velocity =v=18km/hr=18x5/18=5 m/stime= 2secAverage acceleration =a=?Formula to be used : a= v-u/ta=5-0/2 =2.5 m/s²∴ Magnitude of his average acceleration is 2.5m/s²
```
3 months ago
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