Question 11:
easyAt \(n^{\text{th}}\) 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 :
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).
Question 12:
moderateA car is travelling at \(72\text{ kmh}^{-1}\) and is \(20\text{ m}\) from a barrier when the driver puts on the brakes. The car hits the barrier 2s later. What is the magnitude of the constant deceleration?
Initial velocity \(u = 72\text{ km/h} = 20\text{ m/s}\). Using \(s = ut - \frac{1}{2}at^2\): \(20 = 20(2) - \frac{1}{2}a(2)^2 ⇒ 20 = 40 - 2a ⇒a = 10\text{ ms}^{-2}\).
Question 13:
easyA 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
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, \(s' = 4 \times 6 = 24\text{ m}\).
Question 14:
easyA train started from rest from a station and accelerated at \(2\text{ ms}^{-2}\) for 10 s. Then, it ran at constant speed for 30 s and thereafter it decelerated at \(4\text{ ms}^{-2}\) until it stopped at the next station. The distance between two station is
Max velocity \( v = 20 \text{ ms}^{-1}\). Acceleration distance \(s_1 = \frac{1}{2}(2)(10^2) = 100\text{ m}\). Constant velocity distance \(s_2 = 20 \times 30 = 600\text{ m}\). Deceleration distance \(s_3 = \frac{20^2}{2 \times 4} = 50\text{ m}\).
Total distance = 750 m.
Question 15:
easyA particle moving with a velocity equal to 0.4 m/s is subjected to an uniform acceleration of
\( 0.15 m/s^{2} \) for 2 sec in a direction at right angles to its initial direction of motion. The resultant velocity is :
Initial velocity along x-axis is \(v_x = 0.4\text{ m/s}\). Velocity developed along the perpendicular y-axis is \(v_y = a_y t = 0.15 \times 2 = 0.3\text{ m/s}\). Resultant velocity is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{0.4^2 + 0.3^2} = 0.5\text{ m/s}\).
Question 16:
difficultA 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 :
For uniform acceleration, the velocity midway between two points is given by the formula: \(v_{mid} = \sqrt{\frac{v_1^2 + v_2^2}{2}} = \sqrt{\frac{30^2 + 40^2}{2}} = \sqrt{\frac{2500}{2}} = 25\sqrt{2}\text{ km/hr}\).
Question 17:
easyA 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:
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 \( v = 2\sqrt{gs \sin\theta} \).
Question 18:
easyIf 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:
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 \(s = \frac{20+60}{2} \times 5 = 200\text{ m}\).
Question 19:
difficultA 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 \(g = 10\text{ ms}^{-2}\)):
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 \(y = \frac{5H}{6}\).
Question 20:
easyA 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
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 \(1 : 5\).