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3-2.Motion in Plane
medium
A car is moving at a speed of $40 \,m / s$ on a circular track of radius $400 \,m$. This speed is increasing at the rate of $3 \,m / s ^2$. The acceleration of car is ....... $m / s ^2$
A$4$
B$7$
C$5$
D$3$
Solution
(c)
$v=40 \,ms ^{-1}$
$r=400 \,m$
$a_T=3 \,ms ^{-2}$
$a_c=\frac{V^2}{r}=\frac{40 \times 40}{400}=4 \,ms ^{-2}$
$a=\sqrt{a_C^2+a_T^2}$
$a=\sqrt{4^2+3^2}=5 \,ms ^{-2}$
$a=5 \,ms ^{-2}$
$v=40 \,ms ^{-1}$
$r=400 \,m$
$a_T=3 \,ms ^{-2}$
$a_c=\frac{V^2}{r}=\frac{40 \times 40}{400}=4 \,ms ^{-2}$
$a=\sqrt{a_C^2+a_T^2}$
$a=\sqrt{4^2+3^2}=5 \,ms ^{-2}$
$a=5 \,ms ^{-2}$
Standard 11
Physics
