Question

An airplane reaches its takeoff speed of 60 m/s with constant acceleration in 30 s, starting from rest. The time it takes to go from 40 m/s to 60 m / s is: A car starts moving from rest with an acceleration of 0.40 m/ s2. A minute later the acceleration stops, and the car moves with constant speed for 5.0 minutes. The car's brakes are finally applied and stops in 2.0 minutes slowing evenly. The total distance traveled by the car is: A child drops a stone into a well and hears the splash of the stone hitting the water 5.3 s later. If the speed of sound from traveling over the water level to its surface is 330 m/s, how deep is the water from under the ground?

Answer 1

The question are based on kinematics and hence we will use the newton's equations to solve these. Also we have to take care of the units. The detailed answer is given below :

Sues. 1 Airplane reaches its speed (60m/sec) in 30 sec. so the acceleration = a= AV Goo boo am sec² 30 time required to go from 40 m/sec to com/sec. use o= ut at v= final velocity= 60 m/s2 = initial velocity = 40 m/see 60 = 40 + (axt) 60-40 = 9 20=2+ [t= 10 sec 9-2 - The car starts moving from rest, u=0 a aceleration = 0.40 m/sec The acceleration stops aqur Iminute, t= Imin= 60 sec distance traveleda S = ut & ļ at 2 15.0 + 4 x 0.40x60173 790m 1 0 The final speed at the end of 1 min! u= ut at = 0+ (0.40) x 60 = 94m/sec Now for 5 minutes, speed is constant, so the distance travelled :- u=$ » S ot

so S = 24 m x (5 x 60) se BC In 190 Isa 72oom.] next two minutes, the car comes to rest. va omisec u= 24 misec, and to gmin= 12osee so u= ut at 0 : 84 + ax 180) .-24= axigo as -84. -0.2 misec? So distance travelled! Si utt I at? - (84 x 190 ) + (-0.8)x(120) = 9880 - 1440 is = 1440 m ? total distance travelled Stot = 720 +7900 + 1440 r Stot = 9360 ml

The depth = dmeks Now the " question question lould be divided into the parts.- The stone travels the water surfene free fall) The sound reaches to listener. Now under hee falli- da ta g ti? > d=1x9.88 t ² = 4.9 +² -① distance travelled by sound! da o tz = 330xt2= 330 tz - total time taken . tittra 5-3 sec Itaa 5-3.-t from O&G 4.9 +²=. 330 tz 4.9 ti²= 330 (5-3-ti) 4.9 t? + 330 t, - 1749 = 0 sowe queadratic eon's ti= -b I 1/6²you => z= -330 À 11/3301²- (4x4.942-1949) ga 2x 4.9 - -330 | 143,186 - 2x4.9 t- -330 + 378.39 | non from eq. 2x4.9 take the only I d= 4.9 ti? t= -330 + 378.39 1 = 4,9 x(4.937 12 - 2x4.9 d: 119.43am - 48.391 9.8 I to = 4.93794

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