Two cars started off the race with accelerations 3 m/ s² and 8 m/ s². Find the time after which the separation between them is 64 km
S=ut+1/2at^2
X=0+1/2*3*t^2------- (1)
64000+X=0+1/2*8*t^2-------(2) because the distance between the two cars is 64km
On simplification of 1 and 2,
t^2= 25600
t= 160s
To find the time after which the separation between the two cars is 64 km, we can use the equation of motion:
s = ut + (1/2)at²
where:
s = separation
u = initial velocity
t = time
a = acceleration
For the first car:
acceleration, a₁ = 3 m/s²
For the second car:
acceleration, a₂ = 8 m/s²
Since both cars start from rest:
Initial velocity, u₁ = 0 m/s for the first car
Initial velocity, u₂ = 0 m/s for the second car
Let's calculate the time required for each car to reach a separation of 64 km:
First car:
Separation, s₁ = 64 km = 64,000 m
Acceleration, a₁ = 3 m/s²
Initial velocity, u₁ = 0 m/s
64,000 = 0 + (1/2)(3)t₁²
64,000 = 1.5t₁²
t₁² = 64,000 / 1.5
t₁² = 42,666.67
Taking the square root of both sides:
t₁ = √(42,666.67)
t₁ ≈ 206.65 seconds
Second car:
Separation, s₂ = 64 km = 64,000 m
Acceleration, a₂ = 8 m/s²
Initial velocity, u₂ = 0 m/s
64,000 = 0 + (1/2)(8)t₂²
64,000 = 4t₂²
t₂² = 64,000 / 4
t₂² = 16,000
Taking the square root of both sides:
t₂ = √(16,000)
t₂ ≈ 126.49 seconds
Therefore, the time after which the separation between the two cars is 64 km is approximately 206.65 seconds for the first car and approximately 126.49 seconds for the second car.
To find the time after which the separation between the two cars is 64 km, we can use the equations of motion.
Let's assume that the initial separation between the two cars is zero, meaning they started side by side. Let the time be denoted by 't', the acceleration of the first car be 'a1' (3 m/s²), and the acceleration of the second car be 'a2' (8 m/s²).
The equation of motion for the first car is:
s1 = u1*t + (1/2)*a1*t²
The equation of motion for the second car is:
s2 = u2*t + (1/2)*a2*t²
Since they started off side by side, the initial positions of both cars will be zero:
s1 = 0 and s2 = 0
Substituting the values into the equations of motion:
0 = 0*t + (1/2)*3*t² ==> (1/2)*3*t² = 0 ==> t² = 0 ==> t = 0
0 = 0*t + (1/2)*8*t² ==> (1/2)*8*t² = 0 ==> t² = 0 ==> t = 0
From the equations of motion, we can see that both cars will stay side by side for the entire time, as the initial positions and velocities are the same.
Therefore, the time after which the separation between the two cars is 64 km is never, as they will always be side by side.
Why is it per second squared?
Distance = rate * time
64 = (8-3)/1000 * time
Solve for time.