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A particle starts from rest, accelerates at $2 \mathrm{~m} / \mathrm{s}^2$ for $10 \mathrm{~s}$ and then goes for constant speed for $30 \mathrm{~s}$ and then decelerates at $4 \mathrm{~m} / \mathrm{s}^2$ till it stops. What is the distance travelled by it
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The correct answer is:
750 m
Velocity acquired by body in $10 \mathrm{sec}$
$v=0+2 \times 10=20 \mathrm{~m} / \mathrm{s}$
and distance travelled by it in $10 \mathrm{sec}$
$S_1=\frac{1}{2} \times 2 \times(10)^2=100 \mathrm{~m}$
then it moves with constant velocity $(20 \mathrm{~m} / \mathrm{s}) \quad$ for $30 \mathrm{sec}$
$S_2=20 \times 30=600 \mathrm{~m}$
After that due to retardation $\left(4 \mathrm{~m} / \mathrm{s}^2\right)$ it stops
$S_3=\frac{v^2}{2 a}=\frac{(20)^2}{2 \times 4}=50 \mathrm{~m}$
\(\Rightarrow\) Total distance travelled by it is \(100+600+50=750\)
$v=0+2 \times 10=20 \mathrm{~m} / \mathrm{s}$
and distance travelled by it in $10 \mathrm{sec}$
$S_1=\frac{1}{2} \times 2 \times(10)^2=100 \mathrm{~m}$
then it moves with constant velocity $(20 \mathrm{~m} / \mathrm{s}) \quad$ for $30 \mathrm{sec}$
$S_2=20 \times 30=600 \mathrm{~m}$
After that due to retardation $\left(4 \mathrm{~m} / \mathrm{s}^2\right)$ it stops
$S_3=\frac{v^2}{2 a}=\frac{(20)^2}{2 \times 4}=50 \mathrm{~m}$
\(\Rightarrow\) Total distance travelled by it is \(100+600+50=750\)
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