Two motorcycles are at rest and are separated by 24 ft they start at the same time and same direction , the one in the back is having acceleration 3 ft/ sec^2 and the one in front is 2 ft/sec^2 can be machine at the back overtake in front?

A. How long thus it takes to overtake?
B. How far this is goes before it cathes up?
C. How fast its machine going at the same time on overtaking?

a. d1=.5 (2)t^2

d2=.5 (3)t^2 +24
but d1=d2 then
put d1, d2 in and solve for time t.
b. use the d2 equation
c.v2=3*t^2

To determine if the motorcycle at the back can overtake the motorcycle in front, we need to compare their positions and velocities at a given time. Let's break down each part of the question:

A. How long does it take to overtake?

To find the time it takes for the motorcycle at the back to overtake the one in front, we can use equations of motion.

Let's consider the motorcycle in front as Motorcycle A and the one at the back as Motorcycle B.

For Motorcycle A:
Initial velocity (u) = 0 ft/sec (as it starts from rest)
Acceleration (a) = 2 ft/sec^2
Distance covered (s) = 24 ft

Using the equation:
s = ut + 0.5at^2

Rearranging the equation to find time (t):
t = sqrt(2s/a)

t = sqrt(2 * 24 / 2)
t = sqrt(24) ≈ 4.9 seconds

Therefore, it takes approximately 4.9 seconds for Motorcycle B to overtake Motorcycle A.

B. How far does it go before it catches up?

To find the distance Motorcycle B covers before overtaking Motorcycle A, we need to calculate the distance traveled by both motorcycles at the time of overtaking.

For Motorcycle A:
Distance covered (s) = 24 ft

For Motorcycle B:
Initial velocity (u) = 0 ft/sec (as it starts from rest)
Acceleration (a) = 3 ft/sec^2
Time (t) = 4.9 seconds (from Part A)

Using the equation:
s = ut + 0.5at^2

For Motorcycle B:
s = 0 * 4.9 + 0.5 * 3 * (4.9)^2
s ≈ 35.81 ft

Therefore, Motorcycle B will travel approximately 35.81 ft before catching up to Motorcycle A.

C. How fast is Motorcycle B going at the time of overtaking?

To find the velocity of Motorcycle B at the time of overtaking, we can use the equation of motion:

v = u + at

For Motorcycle B:
Initial velocity (u) = 0 ft/sec (as it starts from rest)
Acceleration (a) = 3 ft/sec^2
Time (t) = 4.9 seconds (from Part A)

v = 0 + 3 * 4.9
v ≈ 14.7 ft/sec

Therefore, Motorcycle B is traveling at approximately 14.7 ft/sec at the time of overtaking.