At \(n^{\text{th}}\) second of the motion, the distance moved by the body is 3 times the distance moved in the previous second. The motion is uniformly accelerated & started from rest. The value of (n) is :
1. (3)
2. (2)
3. (1)
4. (4)
View Answer
Distance in \( n^{\text{th}} \) second is \( S_n = u + frac {a}{2}(2n-1)) \). Starting from rest \( u=0 , S_n =frac {a}{2}(2n-1).\) Given
\( S_n = 3 S_{n-1} \), we have 2n-1 = 3(2n-3). Solving gives (4n = 8), hence (n = 2).
A car moving with a speed of \(50\text{ kmh}^{-1}\), can be stopped by brakes after at least \(6\text{ m}\). If the same car is moving at a speed of \(100\text{ kmh}^{-1}\), the minimum stopping distance is
1. \(12\text{ m}\)
2. \(18\text{ m}\)
3. \(24\text{ m}\)
4. \(6\text{ m}\)
View Answer
Stopping distance \(s\) is proportional to \(u^2\). When the speed is doubled (from 50 to 100), the stopping distance increases by a factor of 4. Thus, \(s' = 4 \times 6 = 24\text{ m}\).
A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/hr and 40 km/hr respectively. The velocity of the car midway between P and Q is :
1. 33.3 km/hr
2. \(20\sqrt{3} km / hr\)
3. /(25\sqrt{2} km / hr/)
4. 35 km/hr
View Answer
For uniform acceleration, the velocity midway between two points is given by the formula: \(v_{mid} = \sqrt{\frac{v_1^2 + v_2^2}{2}} = \sqrt{\frac{30^2 + 40^2}{2}} = \sqrt{\frac{2500}{2}} = 25\sqrt{2}\text{ km/hr}\).
A body of mass \( m \) is projected along a rough inclined plane (having an angle of inclination with horizontal \( \theta \), equal to angle of repose) with a velocity \( v \). It travels up a maximum distance \( s \) before it comes to a halt. Then \( v \) is:
1. \( \sqrt{gs \cos\theta} \)
2. \( 2\sqrt{gs \sin\theta} \)
3. \( 2\sqrt{gs \tan\theta} \)
4. \( \sqrt{gs \tan^2\theta} \)
View Answer
Since the inclination equals the angle of repose, \( \mu = \tan\theta \). The acceleration down the incline during upward motion is \( a = g \sin\theta + \mu g \cos\theta = 2g \sin\theta \). Using \( v^2 = 2as \), we get \( v = 2\sqrt{gs \sin\theta} \).
If velocity of a car increases uniformly from \(20\text{ m/sec}\) to \(60\text{ m/sec}\) in a time interval of 5 seconds. Then distance travelled during this interval is:
1. 120 m
2. 180 m
3. 200 m
4. 400 m
View Answer
Since acceleration is uniform, the distance is given by the formula \(s = \frac{u+v}{2} \times t\). Substituting \(u=20\text{ m/s}\), \(v=60\text{ m/s}\), and \(t=5\text{ s}\), we get \(s = \frac{20+60}{2} \times 5 = 200\text{ m}\).
A paratrooper jumps from a height \(H\). The parachute can provide a uniform deceleration of \(2\text{ ms}^{-2}\). The height above the ground at which the parachute should be opened so that he touches ground with zero speed is (take \(g = 10\text{ ms}^{-2}\)):
1. \(\frac{H}{6}\)
2. \(\frac{4H}{5}\)
3. \(\frac{5H}{6}\)
4. \(\frac{6H}{7}\)
View Answer
Let \(h\) be free fall and \(y\) be decelerating height, so \(H = h + y\). Speed before parachute opens: \(v^2 = 2gh = 20h\). Deceleration phase: \(0 = v^2 - 2ay = v^2 - 4y\), which gives \(20h = 4y \Rightarrow y = 5h\). Since \(H = 6h\), we find \(y = \frac{5H}{6}\).
A car starts from rest and moves with constant acceleration \(a\). The ratio of distances covered in the first second to the distance covered in third second is
1. \(1 : 2\)
2. \(2 : 1\)
3. \(3 : 1\)
4. \(1 : 5\)
View Answer
The distance covered in the \(n\)-th second from rest is given by \(S_n = frac{a}{2}(2n - 1)\). For the first second (\(n=1\)), \(S_1 = frac{a}{2}\), and for the third second (\(n=3\)), \(S_3 = frac{5a}{2}\). The ratio of \(S_1 : S_3\) is \(1 : 5\).