362430
In a Shopping Mall, a person walks up a stalled escalator in \(90\,s\). When standing on the same escalator, now moving, he is carried in\(60\,s\). The time he would take to walk up the moving escalator will be
1 \(27\,s\)
2 \(72\,s\)
3 \(18\,s\)
4 \(36\,s\)
Explanation:
\({t_{1}=90 s}\) \(\Rightarrow 90=\dfrac{d}{v_{m}}\)\(w.r.t.\)speed of man \({=\dfrac{d}{90}}\) \({t_{2}=60 s}\) \({\Rightarrow 60=\dfrac{d}{v_{s}}}\) speed of escalator, \({v_{s}=\dfrac{d}{60}}\) \({t=\dfrac{d}{v_{m}+v_{s}}=\dfrac{d}{d / 90+d / 60}}\) \({=\dfrac{60 \times 90}{150}=36 {~s}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362431
A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?
1 \(x < 0, v < 0, a>0\)
2 \(x>0, v < 0, a < 0\)
3 \(x>0, v < 0, a>0\)
4 \(x>0, v>0, a < 0\)
Explanation:
As the lift is coming in downward direction, displacement will be negative. We have to see whether the motion is accelerating or retarding. We know that due to downward motion displacement will be negative. When the lift reaches \(4^{\text {th }}\) floor is about to stop hence, motion is retarding in nature. Hence \(x < 0 ; a>0\). As displacement is in negative direction, velocity will also be negative i.e., \(v < 0\).
362430
In a Shopping Mall, a person walks up a stalled escalator in \(90\,s\). When standing on the same escalator, now moving, he is carried in\(60\,s\). The time he would take to walk up the moving escalator will be
1 \(27\,s\)
2 \(72\,s\)
3 \(18\,s\)
4 \(36\,s\)
Explanation:
\({t_{1}=90 s}\) \(\Rightarrow 90=\dfrac{d}{v_{m}}\)\(w.r.t.\)speed of man \({=\dfrac{d}{90}}\) \({t_{2}=60 s}\) \({\Rightarrow 60=\dfrac{d}{v_{s}}}\) speed of escalator, \({v_{s}=\dfrac{d}{60}}\) \({t=\dfrac{d}{v_{m}+v_{s}}=\dfrac{d}{d / 90+d / 60}}\) \({=\dfrac{60 \times 90}{150}=36 {~s}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362431
A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?
1 \(x < 0, v < 0, a>0\)
2 \(x>0, v < 0, a < 0\)
3 \(x>0, v < 0, a>0\)
4 \(x>0, v>0, a < 0\)
Explanation:
As the lift is coming in downward direction, displacement will be negative. We have to see whether the motion is accelerating or retarding. We know that due to downward motion displacement will be negative. When the lift reaches \(4^{\text {th }}\) floor is about to stop hence, motion is retarding in nature. Hence \(x < 0 ; a>0\). As displacement is in negative direction, velocity will also be negative i.e., \(v < 0\).