Question

A car start running from rest under uniform acceleration of 2.5 m/s2 for 8 second. The final velocity of car will be​

Answer 1

Answer:

Explanation:

Avg speed × total times = total distance

20×25=500

Since acceleration in the 1st interval is same as deceleration in the 3rd interval and deceleration is from the same velocity achieved after acceleration, t  

1

​

=t  

3

​

 

s  

1

​

=  

2

1

​

×5t  

1

2

​

 

s  

2

​

=(5t  

1

​

)t  

2

​

 

s  

3

​

=(5t  

1

​

)×t  

3

​

−  

2

1

​

×5×t  

1

2

​

=(5t  

1

​

)×t  

1

​

−  

2

1

​

×5×t  

1

2

​

 

s  

1

​

+s  

2

​

+s  

3

​

=5t  

1

​

t  

2

​

+5t  

1

2

​

 

⇒5t  

1

2

​

+5t  

1

​

t  

2

​

=500−−−−1

2t  

1

​

+t  

2

​

=25−−−−2

Solving (1 ) and   (2) ⇒t  

1

​

=5,t  

2

​

=15sec

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