Question

a car is moving with a speed of 30m/s on a circular path of radius 500m.its speed is increasing at the rate of 2m/s.what is acceleration​

Answer 1

Answer:

Given:

Linear speed = 30 m/s

Tangential acceleration = 2 m/s²

To find:

Net acceleration of the object

Concept:

First you have to calculate the centripetal acceleration. Then you have to find the net acceleration by vector addition.

Calculation:

Centripetal acc. = v²/R

=> Centripetal acc. = (30)²/500

=> Centripetal acc. = 900/500

=> Centripetal acc. = 9/5

=> Centripetal acc = 1.8 m/s².....(c-acc.)

Again Tangential acceleration (as given in the question ) = 2 m/s²..... (t-acc.)

Net acceleration of the object

= √{(c-acc.)² + (t-acc.)²}

= √{(1.8)² + (2)²}

= √{3.24 + 4}

= √7.24

= 2.690 m/s²

So the final answer is 2.69 m/s²

Answer 2

$\Huge{\underline{\underline{\red{\sf{Answer :}}}}}$

$\large \tt \: Given\begin{cases} \sf{Linear \: speed \: (v) \: = \: 30 \: m {s}^{ - 1} } \\ \sf{Radius \: = \: 500 \: m} \\ \sf{Linear \: Accelertion \: = \: 2 \: m {s}^{ - 2} } \end{cases}$

Solution :

As we know Formula for Centripital (α) Acceleration is :

$\Large \displaystyle {\underline{\boxed{\sf{\alpha \: = \: \frac{v^2}{R}}}}}$

Put Values

⇒ α = (30)²/500

⇒α = 900/500

⇒α = 9/5

⇒α = 1.8 m/s²

$\large {\underline{\boxed{\red{\sf{\alpha \: = \: 1.8 \: ms^{-2}}}}}}$

∴ Centripital acceleration is 1.8 m/s²

__________________________

Now,

we have to find net acceleration.

So,

$\Large \leadsto {\sf{a_{net} \: = \: \sqrt{ (a_{centripital}) ^2 \: + \: (a_{linear})^2}}}$

Put Values

⇒a(net) = √(1.8)² + (2)²

⇒a(net) = √3.24 + 4

⇒a(net) = √7.24

⇒a(net) = 2.690

$\Large {\underline{\boxed{\sf{a_{net} \: = \: 2.690 \: ms^{-2}}}}}$

∴ Net acceleration is 2.69 m/s²

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