Kinematics physics questions?

Question

A police car at rest, passed by a speeder traveling at a constant 110 km/h, takes off in hot pursuit. The police officer catches up to the speeder in 700 m, maintaining a constant acceleration.

a) Calculate the required police car acceleration.

b) Calculate the speed of the police car at the overtaking point.

A car is behind a truck going 25 m/s on the highway. The driver looks for an opportunity to pass, guessing that his car can accelerates at 1.0 m/s^2, and he gauges that he has to cover the 20 m length of the truck, plus 10 m clear room at the rear of the truck and 10 m more at the front of it. In the oncoming lane, he sees a car approaching, probably also traveling at 25 m/s. He estimates that the car is about 400 m away. Should he attempt the pass? Details and explain.

Answer 1

A police car at rest, passed by a speeder traveling at a constant 110 km/h, takes off in hot pursuit. The police officer catches up to the speeder in 700 m, maintaining a constant acceleration.

<< a) Calculate the required police car acceleration. >>

Let

T = time required for the speeder to negotiate 700 m (or 0.7 km)

T = (0.70/110) * 3600 = 23 seconds

This is also the time that the police car will take to catch up with the speeder.

Let

a = acceleration of the police car to catch up the the speeder

Working formula is

S = VT + (1/2)aT^2

where

S = 700 meters

V = initial velocity = 0 (since police car started from rest)

T = time = 23 sec.

a = acceleration

Substituting appropriate values,

700 = 0(23) + (1/2)(a)(23^2)

Solving for "a",

a = 2(700)/23^2

a = 2.65 m/sec^2

<< Calculate the speed of the police car at the overtaking point.>>

Formula to use is

Vf^2 - V^2 = 2aS

where

Vf = velocity of police car at the overtaking point

and all the other terms have been previously defined.

Substituting values,

Vf^2 - 0 = 2(2.65)(700)

Vf^2 = 3710

Vf = 60.91 m/sec = 219.28 km/hr.

<< A car is behind a truck going 25 m/s on the highway. The driver looks for an opportunity to pass, guessing that his car can accelerates at 1.0 m/s^2, and he gauges that he has to cover the 20 m length of the truck, plus 10 m clear room at the rear of the truck and 10 m more at the front of it. In the oncoming lane, he sees a car approaching, probably also traveling at 25 m/s. He estimates that the car is about 400 m away. Should he attempt the pass? Details and explain >>

The total distance that the car has to negotiate during the overtaking process is

D = 40 + 25T

where

40 = 10 + 20 + 10 meters

25 = speed of the truck (since the truck will not stop during the process, it will still travel a certain distance while the car is in the process of overtaking it).

T = time it will take to overtake the truck

The working equation is

D = VT + (1/2)aT^2

where

D = 40 + 25T

V = initial velocity of the car = 25 m/sec (same as that of the truck)

a = car's acceleration = 1 m/sec^2

T = time for car to overtake the truck

Substituting appropriate values,

40 + 25T = 25T + (1/2)(1)T^2

Solving for "T"

T = sqrt (2*40)

T = 8.94 seconds

This is the time (T = 8.94 sec) required for the car to overtake the truck at the given conditions.

At T = 8.94 sec., the total distance travelled by the car will be,

D = 40 + 25(8.94)

D = 263.5 meters

Initially, the oncoming car is 400 meters away and estimated to be running at 25 m/sec also. To determine whether it is safe to pass or not, the distance travelled by the oncoming car must be less than

400 - 263.5 meters OR 136.5 meters in the same time frame of 8.94 seconds.

So, calculating the distance travelled by the oncoming car in 8.94 sec.,

Distance = 25 * 8.94 = 223.5 meters

Since the calculated distance travelled by the oncoming car (within the same time frame of 8.94 sec) is 223.5 meters (and greater than the 136.5 meters), then the car, running behind the truck, must not attempt to pass.