An ambulance is currently traveling at 15m/s, and is accelerating with a constant acceleration of 5
m/s2. 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?
Answers
Given:
- Initial velocity, u₁ = 15 m/s
- Terminal velocity, u₂ = 30 m/s
- Acceleration, a = 5 m/s²
Solution:
First we find the time required to travel with the velocity 30 m/s.
We use the formula: u₂ = u₁ + at, where t is the time required
- ⇒ 30 = 15 + 5t
- ⇒ 5t = 15
- ⇒ t = 3 s
Now we find the average velocity.
If v be the average velocity, then
- v = (u₁ + u₂)/2
- ⇒ v = (15 + 30)/2 m/s
- ⇒ v = 45/2 m/s
- ⇒ v = 22.5 m/s
Now we have to find the distance travelled by the ambulance to match the car's velocity.
We use the formula: s = vt, where s is the required distance to be travelled
- ⇒ s = 22.5 * 3 m
- ⇒ s = 67.5 m
∴ the ambulance must travel 67.5 m until it matches the car's velocity.
The ambulance travels 67.5 meters until it matches the car's velocity.
Given:
Initial velocity = u₁ = 15 m/s
Terminal velocity = u₂ = 30 m/s
Acceleration = a = 5 m/s²
Explanation:
The equation of motion need to be used in this problem is:
u₂ = u₁ + at
On substituting the values, we get,
30 = 15 + (5 × t)
15 = 5 × t
∴ t = 3 s
Now, the average velocity is given by the formula:
v = (u₁ + u₂)/2
On substituting the values, we get,
v = (15 + 30)/2
v = 45/2
∴ v = 22.5 m/s
Now, the distance traveled is given by the formula:
s = v × t
On substituting the values, we get,
s = 22.5 × 3
∴ s = 67.5 m