a bus running at a speed of 18 km/h is stopped in 2.5 seconds by applying brakes. calculate the retardation produced.
u (initial velocity) = 18 km/hr 
t ( time )= 2.5 sec
v( final velocity) = 0 m/s
from, the first equation of motionv = u +  a×t0 = 5 +  a×2.5a = - 2 m/s2
miles/hour is just that.
 
(3/4)/(1/12) = (3/4)(12) = 9 mph the distance covered by a car in 5 h if it is moving with a speed of 75 km/h is\_\_\_\_\_\_\_\_? let the speed of train = x km /hlet the speed of car = y km/hcase 1 : 120 km by train and 480 by cartime taken to travel 120 km by train = 120x hrstime taken to travel 120 km by car = 480x hrsaccording to first condition :120x + 480y = 8⇒1x + 4y = 115     
1case 2 : 200 km by train and 400 by cartime taken to travel 200 km by train = 200x hrstime taken to travel 400 km by car = 400x hrsaccording to second condition :    200x + 400y = 82060 = 813 = 253         as, 20 min = 2060 hr = 13 hr⇒1x + 2y = 124     
2subtracting 2 from 1, we get      2y = 115 - 124⇒2y = 8-5120⇒y = 80now, from 1, we get    1x + 480 = 115⇒1x = 11 - 120 = 160⇒x = 60so, speed of train = 60 km/hspeed of car = 80 km/h a bob of mass m is suspended by inextensible string of length l from a fixed point. the bob is given a speed of at the bottom point. find the tension in the string when string deflects through an angle 120 degree from vertical. try doing a compression test hint: here, we will proceed by finding out the relative speed between the two cars and then using the formula i.e., time taken$ = \dfrac{{{\text{distance travelled}}}}{{{\text{speed}}}}$ to determine after how many hours these cars will meet each other.complete step-by-step answer:given, distance between the two cars d=160 kmstarting time for both the cars when they left their respective initial positions a and b is 8:10 amspeed of the first car, x=50 km per hourspeed of the second car, y=30 km per hourin order to obtain the relative speed between these two cars which are moving towards each other (i.e., in opposite directions), the individual speeds of these cars will be added.so, the relative speed between the two given cars, r= x+y=50+30=80 km per houras we know that, time taken$ = \dfrac{{{\text{distance travelled}}}}{{{\text{speed}}}}$so, time when both the cars will meet=$\dfrac{{{\text{initial distance between both the cars}}}}{{{\text{relative speed between the cars}}}} = \dfrac{{\text{d}}}{{\text{r}}} = \dfrac{{160}}{{80}} = 2$ hours.after 8:10 am, both the cars which are initially at different places a and b will meet each other in the next 2 hours.therefore, time when they will meet each = 8:10 am + 2 hours = 10:10 amhence, option a is correct.note: in this particular problem, since the given two cars are moving in opposite directions (or moving towards each other) that’s why the individual speed of the cars are added in order to determine the relative speed between them but if they were moving in the same direction then the relative speed between these cars would have been obtained by subtracting the individual speed of these cars. setup the kc expression for the reaction, recall that kc expression will appear as:kc expression for the problem:solving for [n2]:note that ice table is not necessary for this calculation answer is 0.016 m try removeing the full stop in front of range answer: `1.00` -2.8oc