A car is travelling with velocity v in a circular road of radius r .if it is decreasing its speed at the rate of a then the magnitude of resultant acceleration will be?
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The instantaneous acceleration pointing towards the center of the circle is called centripetal acceleration. It is denoted by 'a₍r₎'
It's magnitude is given as
a₍r₎ = v²/r
where 'r' is the radius of curvature.
The tangential component cause a change in its speed and it's magnitude is given as
a₍t₎ = dv/dt = a m/s²
Hence, tangential acceleration is,
a₍t₎ = a m/s²
Now,
Resultant acceleration = a² = a₍r₎² + a₍t₎²
Putting the values , we get:
Resultant acceleration = a² = (v²/r)² + a²
Resultant acceleration = a² = v⁴/r² + a²
Resultant acceleration = a = √ (v⁴/r² + a²)
which is the required answer.
Hopefully it helps.Thanks.
It's magnitude is given as
a₍r₎ = v²/r
where 'r' is the radius of curvature.
The tangential component cause a change in its speed and it's magnitude is given as
a₍t₎ = dv/dt = a m/s²
Hence, tangential acceleration is,
a₍t₎ = a m/s²
Now,
Resultant acceleration = a² = a₍r₎² + a₍t₎²
Putting the values , we get:
Resultant acceleration = a² = (v²/r)² + a²
Resultant acceleration = a² = v⁴/r² + a²
Resultant acceleration = a = √ (v⁴/r² + a²)
which is the required answer.
Hopefully it helps.Thanks.
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