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

2 answers:

8 0

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.

Alex777 [14]4 years ago

7 0

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

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Answer:

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Answer:

Explanation:

given,

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