Question

A car is moving on a circular road r=150m and it's speed is increasing at a rate 8m/s^2. At some instant it's speed is 30m/s, net acceleration of the car at this instant is

Answer 1

Net acceleration must be 8m/s^2 as the speed is increasing at the rate of 8m/s^2.

Answer 2

The net acceleration at that instant is $10\frac{m}{s^{2} }$.

Given,

radius of the circular road, r=150 m

rate of increase of speed i.e, acceleration, a=$8\frac{m}{s^{2} }$

speed at an instant=30 m/s.

To find,

the net acceleration of the car at that instant .

Solution:

Acceleration is the rate of change of velocity.

$a1=\frac{dv}{dt}=8\frac{m}{s^{2} }$$8\frac{m}{s^{2} }$

$F=\frac{mv^{2} }{r} \\ma=\frac{mv^{2} }{r} \\\\a2=\frac{v^{2} }{r} \\a2=\frac{30^{2} }{150} \\a2=\frac{900}{150} \\a2=6\frac{m}{s^{2} }$

a1 and a2 are two components of the net acceleration.

Now, the net acceleration of the car at that instant will be,

$anet=\sqrt{a1^{2}+a2^{2} } \\anet=\sqrt{8^{2}+6^{2} } \\anet=\sqrt{64+36} \\anet=\sqrt{100} \\anet=10 \frac{m}{s^{2} } .$

Net acceleration is $10\frac{m}{s^{2} }$.

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