Calculating Distance with Motorcycle Patrolman and Car

[varsitymsb5]

Please help,

A motorcycle patrolman starts from rest at A 5 seconds after a car, speeding at the constant rate of 124 km/h, passes point A. If the patrolman accelerates at the rate of 6.4 m/s2 until he reaches his maximum permissible speed of 147 km/h, which he maintains, calculate the distance s from point A to the point at which he overtakes the car.

Thanks!

 

[tiny-tim]

welcome to pf!

hi varsitymsb5! welcome to pf!

use the standard constant acceleration equations for each vehicle (you'll need t in one to be t+5 in the other) …

show us what you get! ​

 

[varsitymsb5]

v=vnot+at
147/3.6=6.4t
t=6.38s
s(car)=124/3.6t'
s(moto)=147/3.6(t'-5)

I'm stuck and not sure what to do next.

 

[tiny-tim]

sorry, I'm not following this at all

rewrite the words of the question into equations

 

[alphy]

I would approach this problem by first understanding the given information and identifying the relevant equations and variables. From the given information, we know that the motorcycle patrolman starts from rest and accelerates at a constant rate of 6.4 m/s2 until he reaches his maximum speed of 147 km/h. We also know that the car is traveling at a constant speed of 124 km/h.

To calculate the distance s from point A to the point at which the patrolman overtakes the car, we can use the equation s = ut + 1/2at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Since the patrolman starts from rest, his initial velocity (u) is 0 m/s. We can convert the given speeds of 124 km/h and 147 km/h to m/s by multiplying by 1000/3600, giving us u = 34.4 m/s and v = 40.8 m/s.

Now, we can use the equation v = u + at to find the time it takes for the patrolman to reach his maximum speed:

40.8 m/s = 34.4 m/s + 6.4 m/s^2 * t

t = 1.25 seconds

Next, we can use the equation s = ut + 1/2at^2 to find the distance traveled by the patrolman during this time:

s = 0 m/s * 1.25 seconds + 1/2 * 6.4 m/s^2 * (1.25 seconds)^2

s = 5 meters

This represents the distance traveled by the patrolman before he reaches his maximum speed. Now, we can use the equation s = vt to find the additional distance he travels at his maximum speed of 147 km/h:

s = 40.8 m/s * (t - 1.25 seconds)

s = 40.8 m/s * (5 seconds - 1.25 seconds)

s = 153 meters

Therefore, the total distance traveled by the patrolman from point A to the point at which he overtakes the car is:

s = 5 meters + 153 meters = 158 meters.

In conclusion, using the given information and relevant equations, we can calculate the distance s from point A to the point at which the motorcycle patrolman