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lesantik [10]
3 years ago
6

Which units are used to measure both velocity and speed?

Physics
2 answers:
Stells [14]3 years ago
5 0

                 (Any unit of distance)
divided by
                 (any unit of time)

becomes a unit of speed.  Some favorites are  mile/hour,
meter/second, and  foot/second.

Speed is a part of velocity, so the same units are involved
in velocity.  But velocity needs more than that.

When you have the distance and the time, you know the
speed completely ... there's no more to know about it.

But speed is only part of velocity.  In addition to speed,
there's more to know in order to have a velocity.  That's
the direction of the speed. 
irina1246 [14]3 years ago
4 0
The SI unit for velocity and speed is meters per second
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A ball is gently dropped from a height of 30m. If its velocity increase uniform at the rate of 10m/s with what velocity will it
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Explanation:

<h3>Need to FinD :</h3>

  • The velocity of the ball with strikes the ground.
  • The time of the ball after which it strikes the ground.

\red{\frak{Given}} \begin{cases} & \sf {A\ ball\ is\ gently\ dropped\ from\ a\ height\ of\ {\pmb{\sf{30\ m}}}.} \\ & \sf {Its\ velocity\ increase\ at\ the\ rate\ of\ {\pmb{\sf{10\ m/s^2}}}.} \end{cases}

We know that,

  • The ball is dropped from the height of 30 m as stated in the question. Therefore, it will have a initial velocity of 0 m/s. The distance travelled by the ball will be 30 metres as the distance travelled by the ball is equal to the height of the tower and that is 30 metres.

So, by using the third equation of motion, we will find out the final velocity of the ball.

\rule{200}{3}

Therefore,

\sf \dashrightarrow {2as\ =\ v^2\ -\ u^2} \\ \\ \\ \sf \dashrightarrow {2 \times 10\ m/s^2 \times 30\ m\ =\ v^2\ -\ (0\ m/s)^2} \\ \\ \\ \sf \dashrightarrow {v^2\ =\ 600\ m^2/s^2} \\ \\ \\ \sf \dashrightarrow {v\ =\ \sqrt{600\ m^2s^{-2}}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{Final\ velocity\ (v)\ =\ 24.49\ m/s \approx 24.5\ m/s}}}}_{\sf \blue{\tiny{Required\ velocity\ of\ ball}}}}

∴ Hence, the required final velocity of the ball with which it strikes the ground is 24.5 m/s. Since, we know that the initial velocity of the ball is 0 m/s and the acceleration of the ball is 10 m/s². So, by using the first equation of motion, we will find out the time of the ball after which it stikes the ground.

\rule{200}{3}

Therefore,

\sf \dashrightarrow {v\ =\ u\ +\ at} \\ \\ \\ \sf \dashrightarrow {t\ =\ \dfrac{v\ -\ u}{a}} \\ \\ \\ \sf \dashrightarrow {t\ =\ \dfrac{24.5\ m/s}{10\ m/s^2}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{Time\ (t)\ =\ 2.45\ s}}}}_{\sf \blue{\tiny{Required\ time}}}}

∴ Hence, the required time of the ball after which it strikes the ground is 2.45 seconds.

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3 years ago
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