Circular MotionMediumBloom L3
Question
A car is moving with speed 27 km/h. The driver applies brakes as he approaches a circular turn on the road of radius 80 m and his speed reduces at the constant rate of 0.50 m/s every second. The magnitude of net acceleration is :
Options
A.20 ms–2
B.0.86 ms–2
C.100 ms–2
D.None of these
Solution
Sol. 
Now let magnitude of net acceleration is ‘a’
a = $\sqrt{{a_{c}}^{2} + {a_{t}}^{2}}$
= $\sqrt{(0.7)^{2} + (0.5)^{2}} = .866$ ac = $\frac{v^{2}}{r}$ = $\frac{225}{4 \times 80} = \frac{5}{64}$ = 0.7
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