A motorist drives at 30m/S along a straight road.The speed limit on the road 72km/h.The motorist applies brakes 25m before crossing a speed trap.What must the cars acceleration be to reduce its speed to 72km/h as it's reaches the speed trap?
72000 meters / 3600 seconds = 20 meters/second
so reduce speed from 30 m/s to 20 m/s
average speed during slowdown = 25 m/s
goes 25 meters during that slowdown
so went 25 meters / 25 meters/second = 1 second with the brakes on
change in velocity = -10 m/s
in time 1 second
a = change in velocity / change in time = -10 m/s^2
A motorist drives at 30m. Second on a straight level road.The speed limit is 70 km.h. 50 m before he drives over a speed trap, he slams on brakes.
To find the required acceleration to reduce the speed to 72 km/h, we can follow these steps:
Step 1: Convert the speed limit from km/h to m/s.
Speed limit = 72 km/h
1 km = 1000 m
1 hour = 3600 seconds
72 km/h = (72 * 1000 m) / (3600 s)
= 72,000 m / 3600 s
= 20 m/s
So, the speed limit is 20 m/s.
Step 2: Determine the initial velocity of the car.
Given that the motorist drives at 30 m/s along the road.
Initial velocity (u) = 30 m/s
Step 3: Calculate the change in velocity.
Change in velocity (Δv) = Final velocity (v) - Initial velocity (u)
= 20 m/s - 30 m/s
= -10 m/s (negative sign indicates deceleration)
Step 4: Determine the distance required to achieve the desired change in velocity.
Given that the motorist applies brakes 25 m before the speed trap.
Distance (s) = 25 m
Step 5: Apply the formula to find acceleration (a).
The formula relating acceleration, change in velocity, and distance is:
Δv^2 = 2as
Rearranging the formula to solve for acceleration (a):
a = Δv^2 / (2s)
= (-10 m/s)^2 / (2 * 25 m)
= 100 m^2/s^2 / 50 m
= 2 m/s^2
Therefore, the acceleration required to reduce the speed to 72 km/h as the car reaches the speed trap is 2 m/s^2.
To determine the required acceleration for the car to reduce its speed to 72 km/h as it reaches the speed trap, we need to follow these steps:
Step 1: Convert the speed limit from km/h to m/s.
The speed limit is given as 72 km/h. To convert this to m/s, we need to divide by 3.6 since 1 km/h is equal to 1/3.6 m/s.
Speed limit in m/s = 72 km/h ÷ 3.6 = 20 m/s
Step 2: Calculate the initial velocity of the car.
Given:
Initial velocity (u) = 30 m/s
Final velocity (v) = 20 m/s
Step 3: Calculate the distance covered while decelerating.
Given:
Distance covered (s) = 25 m
Step 4: Apply the formula of motion to find the acceleration.
The formula to calculate acceleration is:
v^2 = u^2 + 2as
Rearranging the formula, we get:
a = (v^2 - u^2) / (2s)
Substituting the values:
a = (20^2 - 30^2) / (2 * 25)
= (400 - 900) / 50
= -500 / 50
= -10 m/s^2
Therefore, the car's acceleration must be -10 m/s^2 (negative because it's decelerating) in order to reduce its speed to 72 km/h as it reaches the speed trap.