An air force jet flying with a speed of 335ms-1 went past an anti-aircraft gun. How far is the aircraft 5s later when the gun was fired?

Salah255

1 year ago

speed is the change in distance per time

in the question , they gave us the parameters for Speed and time , and they are asking us to get distance


the formula for Speed is S=d/t
making distance the subject of formula

the formula become

d=S×T
D=335×5
=1675m

De sage

1 year ago

you don't need to stress yourself with newton's laws of motion ok.

this is just a straightforward question

giantmoleculetutor

9 months ago

v=s/t
s=vt
s=335*5
s=1675m

De sage

1 year ago

hmmm

V=S/t

ahhh
where v= velocity or speed
s= distance
t= time

Sethx3non

3 months ago

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Mavi18Jacob

1 year ago

We will use this first formula to find the acceleration:V=u+at,the answer is 67m/s^2
Then we find the distance with the second formula:S=ut+1/2 at

Tega204

2 years ago

S=(u+v)t/2
=(0+335)5/2
=837.5m

Efranklyn

4 years ago

Since the aircraft is flying horizontal in the sky it is moving with a horizontal component of the velocity look
V = u + gt
but since it is a horizontal movement the gravity does not change and g is assigned 0
Vx = Ux
S = UxT
(Vx)^2= (Ux)^2

Funmiyewa45

7 years ago

V=2d/t