Question

a powerfull motorcycle can accelerate from rest to 28m/s in only 4sec (b) how far does it travel in that time ?
But please give the write solution and answer and the good solution will get brainliest answer

Answer 1

Given, that motorcycle starts from rest

Thus,

u ( Initial velocity ) = 0 m/s

v ( final velocity ) = 28 m/s

t ( time taken ) = 4 s

a ( acceleration ) = v - u / t = 28 - 0 / 4 = 7 m/s^2

s ( distance covered ) = ?

By 2nd equation of motion, we have

s = ut + 1/2 at^2

$s = 0 \times 4 + \frac{1 \times 7 \times 4 \times 4}{2}$

$s = 0 + 56 \\ s = 56$

This distance covered by the motorcycle is 56m

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Answer 2

Answer:

Given, that motorcycle starts from rest

Thus,

u ( Initial velocity ) = 0 m/s

v ( final velocity ) = 28 m/s

t ( time taken ) = 4 s

a ( acceleration ) = v - u / t = 28 - 0 / 4 = 7 m/s^2

s ( distance covered ) = ?

By 2nd equation of motion, we have

s = ut + 1/2 at^2

s = 0 \times 4 + \frac{1 \times 7 \times 4 \times 4}{2}s=0×4+

2

1×7×4×4

\begin{lgathered}s = 0 + 56 \\ s = 56\end{lgathered}

s=0+56

s=56

This distance covered by the motorcycle is 56m

Explanation:

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