Question

A cycle moving at 5m/s is subjected to an acceleration of 0.2m/s2. Find its velocity after 10s. Also calculate the distance covered by it in that time.




​

Answer 1

Answer:

v = u + at

v = 5 + 0.2 × 10

v = 5 + 2

v = 7 m/s

s = ut +1/2at^2

s = 5 × 10 + 1/2 × 0.2 × 10 ^2

s = 50 + 0.1 × 100

s = 50 + 10

s = 60 meters

Explanation:

mark as brainliest please

Answer 2

Given

• u (initial velocity) = 5m/s.
• a (acceleration) = 0.2m/s²
• t (time) = 10 s.

To Find

• v (final velocity) = ?
• s (distance) = ?

Solution

We know that,

$\large \implies \boxed {\boxed {\tt \blue { v = u + at }}}$

Then, velocity will be...

$\tt \implies \: v = u + at \\ \tt \implies \: v = 5 + (0.2)(10) \\ \tt \implies \: v = 5 + 2 \\ \tt \implies \: v = 7m/s$

____________________________

Using formula:

$\\ \large \implies \boxed {\boxed {\tt \blue {v{}^{2} = u{}^{2} + 2as }}}$

Then, distance will be...

$\tt \implies \: v {}^{2} = u {}^{2} + 2as \\ \tt \implies \: (7) {}^{2} = (5) {}^{2} + 2 \times 0.2 \times s \\ \tt \implies \: 49 = 25 + 0.4s \\\tt \implies \: 0.4s = 49 - 25 \\ \tt \implies \:s = \frac{24}{0.4} \\ \tt \implies \:s = 60m$

Final Answers

✏Velocity = 7m/s

✏ Distance = 60m