An ambulance is currently traveling at 15m/s, and is accelerating with a constant acceleration of 5 m/s^2
2 squared . The ambulance is attempting to pass a car which is moving at a constant velocity of 30m/s. How far must the ambulance travel until it matches the car’s velocity?
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Answers
Answer
- Distance covered = 67.5 m
Explanation
Given
- Initial velocity of Ambulance = 15 m/s
- Acceleration = 5 m/s²
- The ambulance is trying to overtake a car
- Velocity of the car = 30 m/s
To Find
- Distance covered by the ambulace untill it reaches the car's velocity?
Solution
So here we may assume the final Velocity of the car as 30 m/s as the ambulance will match the cat's velocity and we shall use the third Equation of motion to get the distance travelled
➝ v²-u² = 2as
- Final Velocity = v = 30 m/s
- Initial Velocity = u = 15 m/s
- Acceleration = a = 5 m/s²
- Distance Covered = s = ?
Substituting the values,
➝ v²-u² = 2as
➝ 30²-15² = 2 × 5 × s
➝ 900-225 = 10s
➝ 675 = 10s
➝ 675/10 = s
➝ Distance = 67.5 m
Therefore the distance covered by the ambulace to reach the velocity of the car is 67.5 m
Answer :—»
Initial velocity, u = 15 m/s
Terminal velocity, v = 30 m/s
Acceleration, a = 5 m/s²
First we find the time required to travel achieve velocity of 30 m/s
We use the formula: v = u + at, where t is the time required
⇒ 30 = 15 + 5t
⇒ 5t = 15
⇒ t = 3 seconds
Now, we find the Average Velocity.
let Vₐᵥ be the Average Velocity, then
Vₐᵥ = (u + v)/2
⇒ Vₐᵥ = (15 + 30)/2 m/s
⇒ Vₐᵥ = 45/2 m/s
⇒ Vₐᵥ = 22.5 m/s
To find the distance travelled by the ambulance to match the car's velocity.
We use the formula:— d = vt
where d is the required distance to be travelled
⇒ d = 22.5 * 3 m
⇒ d = 67.5 m
∴ the ambulance must travel 67.5 m until it matches the car's velocity.
Trick :—
vₙ = velocity at nth second
formula :-
(v₀ + 2v₁............. + 2vₙ₋₁ + vₙ)/2
every term is twice except the initial and terminal.
(15+20+20+25+25+30)/2
=135/2
= 67.5 m