Question

Find the acceleration of a body whose velocity increases from 11ms-1 to 33ms-1 in 10 seconds

Answer 1

Acceleration is equivalent to the rate of change of velocity.

Calculate the difference of velocity, and divide by the total time it took for the change to happen.

$(33 \frac{m}{s} - 11 \frac{m}{s} ) \div 10seconds \\ = 2 \frac{m}{ {s}^{2} }$
That is the answer.

Answer 2

As we know that 
a= dell   V/t
then put values
a=   2ms/^2