a) For the motion of car with uniform velocity we have , $s = ut+\frac{1}{2}at^2$ , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 $m/s^2$

Substituting

$520 = u*223\\ \\u = 2.33 m/s$

The constant velocity of car a = 2.33 m/s

b) We have $s = ut+\frac{1}{2} at^2$

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

$520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2$

Now we have v = u+at, where v is the final velocity

Substituting

v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 $m/s^2$

7 0

3 years ago

A jet aircraft is traveling at 262 m/s in hor-

NeTakaya

Solution :

Speed of the air craft, $S_a$ = 262 m/s

Fuel burns at the rate of, $S_b$ = 3.92 kg/s

Rate at which the engine takes in air, $S_{air}$ = 85.9 kg/s

Speed of the exhaust gas that are ejected relative to the aircraft, $S_{exh}$ =921 m/s

Therefore, the upward thrust of the jet engine is given by

$F=S_{air}(S_{exh}-S_a)+(S_b \times S_{exh})$

F = 85.9(921 - 262) + (3.92 x 921)

= 4862635.79 + 3610.32

= $4.8 \times 10^6 \ N$

Therefore thrust of the jet engine is $4.8 \times 10^6 \ N$ .

3 0

2 years ago