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3 years ago

7

An ambulance is currently traveling at 15m/s, and is accelerating with a constant acceleration of 5 m/s^2. The ambulance is atte

mpting to pass a car that is moving at a constant velocity of 30m/s. How far must the ambulance travel until it matches the car’s velocity?

1 answer:

Lorico [155]3 years ago

5 0

initial velocity=15m/s=u
Acceleration=a=5m/s^2
Final velocity=v=30m/s
Distance be s

Using 3rd equation of kinematics

$\\ \rm\longmapsto v^2-u^2=2as$

$\\ \rm\longmapsto s=\dfrac{v^2-u^2}{2a}$

$\\ \rm\longmapsto s=\dfrac{30^2-15^2}{2(5)}$

$\\ \rm\longmapsto s=\dfrac{900-225}{10}$

$\\ \rm\longmapsto s=\dfrac{675}{10}$

$\\ \rm\longmapsto s=67.5m$

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