
- Speed of the mobile = 250 m/s
- It starts decelerating at a rate of 3 m/s²
- Time travelled = 45s

- Velocity of mobile after 45 seconds

We can solve the above question using the three equations of motion which are:-
- v = u + at
- s = ut + 1/2 at²
- v² = u² + 2as
So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

We are provided with,
- u = 250 m/s
- a = -3 m/s²
- t = 45 s
By using 1st equation of motion,
⇛ v = u + at
⇛ v = 250 + (-3)45
⇛ v = 250 - 135 m/s
⇛ v = 115 m/s
✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
Answer:
<em>1,378.9ms²</em>
Explanation:
Given the following
Distance S = 70.6m
Time t = 0.32secs
Initial velocity = 0m/s
Required
Acceleration
Using the equation of motion
S = ut+1/2at²
Substitute
70.6 = 0+1/2a(0.32)²
70.6 = 0.0512a
a = 70.6/0.0512
a = 1,378.9
<em>Hence the acceleration is 1,378.9ms²</em>
Answer:
(4xy+5ab)(4xy-5ab)
Explanation:
16
-25

4^2 is 16 and 5^2 is 25,
Also, (x-a)(x+a) = x^2-a^2
So, this factorized is:
(4xy+5ab)(4xy-5ab)
Hope this helps!
If the boat is i travling at 10 m/s and the river is 8.0 m/s the boats speed is 18.0 m/s
Answer:
3054.4 km/h
Explanation:
Using the conservation of momentum
momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor
initial momentum = 14900 M km/h
let v be the new speed of the motor so that the
new momentum = 4Mv and the new momentum of the module = M ( v + 94 km/h )
total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M
initial momentum = final momentum
14900 M km/h = 5 Mv + 93M
14900 km/h = 5v + 93
14900 - 93 = 5v
v = 2961.4 km/h
the speed of the module = 2961.4 + 93 = 3054.4 km/h