Question

A bicycle incease in velocity from 10km/h to 15km/h in 6 second calcuate is acceleration

Answer 1

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

Given :

• Final Velocity (v) = 15 km/h = 4.1 m/s
• Initial velocity (u) = 10 km/h = 2.7 m/s
• Time (t) = 6s

_________________________

To Find :

• Acceleration

_________________________

Solution :

We have formula for acceleration :

$\large{\boxed{\sf{a \: = \: \dfrac{v \: - \: u}{t}}}} \\ \\ \longrightarrow \sf{a \: = \: \dfrac{4.1 \: - 2.7}{6}} \\ \\ \longrightarrow {\sf{a \: = \: \dfrac{1.4}{6}}} \\ \\ \longrightarrow {\sf{a \: = \: 0.024}} \\ \\ \longrightarrow {\sf{a \: = \: 2.4 \: \times \: 10^{-2} \: ms^{-2}}}$

Answer 2

Answer:

Given:

Velocity of cycle increases from 10 km/hr to 15 km/hr in 6 seconds.

To find:

Acceleration of the cycle.

Concept:

Acceleration is a vector Quantity representing the change of velocity with respect to time.

$\boxed{ \red{ \large{a = \dfrac{\Delta v}{\Delta t}}}}$

Conversion :

We need to convert the unit of velocity from Km/hr to m/s :

1) 10 km/hr = 10 × (5/18) = 2.77 m/s

2) 15 km/hr = 15 × (5/18) = 4.16 m/s

Calculation:

$\large{a = \dfrac{\Delta v}{\Delta t}}$

$= > a = \dfrac{4.16 - 2.77}{6}$

$= > a = 0.2316 \: m {s}^{ - 2}$

So final answer :

$\boxed{ \red{ \huge{ \bold{ a = 0.2316 \: m {s}^{ - 2} }}}}$

Additional information :

• Acceleration can be broadly classified into 2 types :
• Average acceleration which gives the ratio of velocity change over change of time
• Instantaneous acceleration which gives the instantaneous change of velocity over an infinitesimally small time period.