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solution: In the above question, car is travelling with 5m/s for 10minutes. here he mentioned instantaneous speed ,that means at particular instant how much velocity it has. it has travelled for 10minutes,that means through the journey it is travelling with uniform velocity. so, answer is at t=500s also same speed.
solution: in the question ,the driver managed his car with 5m/s. the time is 10minutes the car is moving with uniform velocity with 5m/s that means ,the car has travelled with 5m/s through out the journey. i.e at 2minutes and 10minutes also ,the car has same velocity. the average speed of the car is same.i.e.5m/s …
solution: car is moving with a speed(u)=20m/s it can stopped by brakes ,that means final velocity (v)=0 in mean time ,it travelled a distance (S)=2m. by given information ,initially it has initial velocity and then comes rest .so, the car is in retarded motion. we have the formula, V2-u2=2aS 0-(20)2 =2a(2) -400=4a a=-100m/S2 The car …
solution: a particle starts from rest (u)=0 the distance covered in first second (t)=1sec we have the formula S=ut+1/2at2 here ,u=0,t=1s,a=a S=0+1/2a(1)2 S1=1/2a the distance covered in 3sec S=ut+1/2at2 S=0+1/2a(3)2 S2=9/2a S1 :S2=1:9 the ratio of the distances covered by body is 1:9
solution: a bus initial velocity u=0 for time =t1 acceleration = α after time t1 ,the velocity will be v1=αt1 this velocity will become initial velocity for retarded motion, u=αt1 then retards with acceleration =-β for time =t2 this is given information , for time t2 ,the body retards v2= αt1-βt2 but v2 is nothing …
solution: initial velocity =u final velocity =u-75% of u =u-3/4u v=u/4 we have the formula v=u+ at retardation= -a u/4=u+(-a)t at=u-u/4 at=3u/4 t=3u/4a if the velocity u of a body moving along a straight line with ,after time 3u/4a the velocity is reduced by 75% of u
solution: In the above problem, acceleration of a particle is (a)=10m/s2 time=5s final velocity=50m/s but it is moving with uniform acceleration. v=u+at 50=u+10*5 u=0m/s i.e it starts with zero initial vvelocity then we have to calculate velocity at 4s for this a=4m/s2 t=4s u=0m/s v=0+10*4 v=40m/s the velocity at 4s is 40m/s.
solution: final velocity:25m/s time :2s a=4m/s2 then, the formula we have v=u+at 25=u+4*2 25-8=u u=17m/s The displacement S=ut+1/2at2 S=17(2)+1/2(4)(2)(2) S=34+8 S=42m The displacement of the body is 42m
solution: u=10m/s a=a v=30m/s S=4km=4000m v2-u2=2as 900-100=2a*4000 800=8000a a=0.1m/s2 then, v=u +at 30=10+0.1*t 30-10=0.1t 20=0.1t t=200s The time taken is 200s.
