Ronil goyal question answers
- 1. Q1. Sol:-Given: Initial velocity=36km/h=36x5/18=10m/s Final velocity=54km/h=54x5/18=15m/s Time =10sec Acceleration=v-u/t =15-10/10=5/10=1/2=0.5 m/s2 Distance =s=? From secondequationof motion: S=ut +1/2 at^2 =10*10+1/2*0.5*10*10 =100+25 =125m So distance travelledis125m Q.2 Sol: A body moving along a straight line at 20 m/s undergoes an acceleration of -4 m/s2. ... A body whose speed is constant (a) must be accelerated (b) might be accelerated (c) has aconstant velocity (d) can not be accelerated. Q3. Sol:The truck istravellingwithainitial velocityof V=54m/s. The truck is travellingwithafinal velocityof u=36m/s. acceleration=V-U/T. giventime =10s. so.36-54/10 =-18/10 =-1.8
- 2. so.the retardationis1.8 negative. Q4. Sol: 1) displacement =0 distance=22/7x20m 2) displacement=20m distance=(1.5)22/7x20 3) displacement=0 distance=(2) 22/7x20m 4) displacement=20m distance=(2.5) 22/7x20m Q5. Sol: Intial velocity =u=10m/s Final velocity =v=20 m/s Time=t=4s Acceleration of the scooter =v-u/t =20-10/4 =10/4 =2.5 m/s² The acceleration produced is 2.5m/s². Q6. Sol: let u=0 as it was in rest a=5ms2 v=u+at v=0+5*5 v=0+25 v=25ms so, velocity of train is 25ms Q8. Sol: Given: Maximum speed= 90km/hr Time=10hrs DISTANCE = 500km
- 3. Average speed= TOTAL DISTANCE/ TOTAL time = 500/ 10 =50km/ hr Ratio of maximum speed and average speed is = 90/ 50 =9:5 Ratio of maximum speed and average speed is =9:5 Q9. Sol: Here , As the car is starting from rest , Initial velocity = u = 0 m/s Final velocity = v = 54 km/hr To convert km/hr to m/s , we multiply it by 5/18 Final velocity = v = 54 km/hr = 54 x 5/18 = 15 m/s Time = t = 2 sec a = acceleration Equation of motion : v = u + at 15 = 0 + 2a
- 4. a = 15/2 = 7.5 m/s² s = distance traveled Equation of motion : s = ut + 1/2 at² = 0 + 1/2 × 7.5 × 2² = 15 metres Q11. Sol: 1) What is the net displacement? Ans. Displacement for the above situation is 0. As we know, that displacement is the shortest path from the initial to the final point. Here, the initial and the final points are the same, and hence, it takes no time to travel. So, the displacement is 0. 2) What is the distance? Ans. Distance for the above situation is 200 m. Here, the ball has to cover a distance of 100 m and while coming back, again it has to cover the same distance, hence, the distance to be covered here is 200 m. 100 + 100 = 200 Q12. Sol: Distance = speed×time d1 = 7.5×2/60 d1 = 0.25 km d2 = 7.5×2/60 d2 = 0.25 km
- 5. Total distance = d1+d2 = 0.5 km Total time = 2+2+56 = 60min = 1 hr Average speed = 0.5/1 = 0.5 km/hr Q13. Sol: distance = 2 km total time = 30 min = 0.5 hourinitial speed = 3km/h for first 1 kmHence time taken = distance/speed = 1/3 h time left = 0.5 - 1/3 = 0.5/3 hdistance left = 1 km Hence speed at which he should walk the second km would be 1/(0.5/3) = 6 km/h. Q14. Sol: v=u+at Here u=10m/s a=2m/s^2 t=2sec v=10+2×2 =10+4 =14m/s Q15. Sol: Initial velocity, u = 20 m/s Distance travelled, s = 5cm=0.05 m Let the retardation produced be ‘a’. The final velocity is v = 0. So, v² = u² +2as => 0 = 20 +2a(0.05) => a = -4000 m/s Thus the retardation/deceleration produced is 4000 m/s² Q10. Sol. First it is in rest so initial speed = u = 0m/s final speed = v = ? acceleration = a = 10m/s² time = t = 5s
- 6. a = (v-u)/t at = v-u at+u = v 10(5) + 0 = v 50m/s = v so it's speed after 5 second is 50m/s