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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?
- a)67.5 km
- b)67.5 m
- c)90 m
- d)45 m
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
An ambulance is currently traveling at 15m/s, and is accelerating with...
Understanding the problem:
The ambulance is moving at 15m/s and accelerating at 5m/s2, while the car is moving at a constant velocity of 30m/s. We need to find out how far the ambulance must travel to match the car's velocity.
Calculating the time taken to match velocities:
To find out how long it takes for the ambulance to match the car's velocity, we can use the formula:
\[ v_f = v_i + at \]
Where:
\( v_f \) = final velocity (30 m/s for the car)
\( v_i \) = initial velocity (15 m/s for the ambulance)
\( a \) = acceleration (5 m/s2)
\( t \) = time taken to match velocities
Rearranging the formula for time:
\( 30 = 15 + 5t \)
\( 5t = 15 \)
\( t = 3 \) seconds
Calculating the distance traveled:
Now that we know it takes 3 seconds for the ambulance to match the car's velocity, we can calculate the distance traveled using the formula:
\[ d = v_i t + \frac{1}{2}at^2 \]
Where:
\( d \) = distance traveled
\( v_i \) = initial velocity (15 m/s)
\( t \) = time taken (3 seconds)
\( a \) = acceleration (5 m/s2)
Substitute the values into the formula:
\[ d = 15(3) + \frac{1}{2}(5)(3)^2 \]
\[ d = 45 + \frac{1}{2}(5)(9) \]
\[ d = 45 + \frac{45}{2} \]
\[ d = 45 + 22.5 \]
\[ d = 67.5 \, meters \]
Therefore, the ambulance must travel 67.5 meters to match the car's velocity. Hence, the correct answer is option B.
The ambulance is moving at 15m/s and accelerating at 5m/s2, while the car is moving at a constant velocity of 30m/s. We need to find out how far the ambulance must travel to match the car's velocity.
Calculating the time taken to match velocities:
To find out how long it takes for the ambulance to match the car's velocity, we can use the formula:
\[ v_f = v_i + at \]
Where:
\( v_f \) = final velocity (30 m/s for the car)
\( v_i \) = initial velocity (15 m/s for the ambulance)
\( a \) = acceleration (5 m/s2)
\( t \) = time taken to match velocities
Rearranging the formula for time:
\( 30 = 15 + 5t \)
\( 5t = 15 \)
\( t = 3 \) seconds
Calculating the distance traveled:
Now that we know it takes 3 seconds for the ambulance to match the car's velocity, we can calculate the distance traveled using the formula:
\[ d = v_i t + \frac{1}{2}at^2 \]
Where:
\( d \) = distance traveled
\( v_i \) = initial velocity (15 m/s)
\( t \) = time taken (3 seconds)
\( a \) = acceleration (5 m/s2)
Substitute the values into the formula:
\[ d = 15(3) + \frac{1}{2}(5)(3)^2 \]
\[ d = 45 + \frac{1}{2}(5)(9) \]
\[ d = 45 + \frac{45}{2} \]
\[ d = 45 + 22.5 \]
\[ d = 67.5 \, meters \]
Therefore, the ambulance must travel 67.5 meters to match the car's velocity. Hence, the correct answer is option B.
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Community Answer
An ambulance is currently traveling at 15m/s, and is accelerating with...
To find displacement with constant acceleration, we can use the formula: d= Δt (va)
Our Δt should simply be the time it takes to go from 15 m/s to 30 m/s at an acceleration of 5 m/s2: Δv = aΔt or 15 = 5Δt, which gives us Δt = 3.
Our va is simply the average velocity, which we can calculate using: va = 1/2(vf + vi) or va = 1/2 (30+15) which gives us va = 22.5 m/s
Using our initial formula, d = Δt (va), d = 3 sec x 22.5 m/s or d = 67.5m
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