A football player, starting from rest at the line of scrimmage, accelerates along a straight line for a time of 1.5 s. Then, during a negligible amount of time, he changes the magnitude of his acceleration to a value of 1.4 m/s2. With this acceleration, he continues in the same direction for another 1.2 s, until he reaches a speed of 3.4 m/s. What is the value of his acceleration (assumed to be constant) during the initial 1.5 s period?
Well,this is a math exercise more than physics.
Compute the value of Vf at the end of each period, and use that as the Vi for the next period.
First period:
Vf=a*1.5
Second Period
Vf=a*1.5+1.4*1.2=3.4
solve the last equation for a.
Thanks, it really helped
To find the value of the football player's acceleration during the initial 1.5 s period, we can use the kinematic equation:
v = u + at
Where:
v = final velocity
u = initial velocity (which is 0 as the player starts from rest)
a = acceleration
t = time
During the initial 1.5 s period, we know that the player's initial velocity is 0 m/s and the final velocity is unknown. So, let's call the final velocity v1. The time taken is 1.5 s. The equation becomes:
v1 = 0 + a(1.5)
Now, during the second period of 1.2 s, the player reaches a speed of 3.4 m/s. Assuming the acceleration is constant, we can use the equation:
v = u + at
Where:
v = final velocity (which is 3.4 m/s)
u = initial velocity (which is the unknown final velocity after 1.5 s, so let's call it v1)
a = acceleration (which is 1.4 m/s^2)
t = time (1.2 s)
3.4 = v1 + (1.4)(1.2)
Now, we have two equations:
v1 = a(1.5) -- Equation (1)
3.4 = v1 + (1.4)(1.2) -- Equation (2)
We can substitute v1 from Equation (1) into Equation (2):
3.4 = [a(1.5)] + (1.4)(1.2)
Now, solve this equation for the acceleration 'a':
3.4 = 1.5a + 1.68
3.4 - 1.68 = 1.5a
1.72 = 1.5a
a = 1.72 / 1.5
a ≈ 1.147 m/s²
Therefore, the value of the player's acceleration during the initial 1.5 s period is approximately 1.147 m/s².
To find the value of the football player's acceleration during the initial 1.5-second period, we can use the equations of motion.
Let's break down the problem into two parts: the first part where the football player accelerates from rest, and the second part where the magnitude of acceleration changes to 1.4 m/s².
In the first part, the football player starts from rest, so his initial velocity (u) is 0 m/s, and the time (t1) is 1.5 s. We need to find the acceleration (a1) during this period.
Using the equation of motion:
v = u + at
Substituting the values we have:
3.4 m/s = 0 m/s + a1 * 1.5 s
Rearranging the equation, we have:
3.4 m/s = 1.5 a1
Now, let's move on to the second part where the magnitude of acceleration changes to 1.4 m/s². The time period for this portion is 1.2 seconds, and we need to find the final velocity.
Using the equation of motion again, with the known values:
v = u + at
The initial velocity (u) will be the final velocity from the first part, which is 3.4 m/s. The time (t2) for this part is 1.2 s, and the acceleration (a2) is 1.4 m/s².
Substituting the values into the equation:
v = 3.4 m/s + 1.4 m/s² * 1.2 s
Solving for v, we get the final velocity:
v = 3.4 m/s + 1.68 m/s
v = 5.08 m/s
Now that we have the final velocity at the end of the 1.2-second period, we can go back to the first part and solve for the acceleration (a1).
Using the equation of motion again, with the known values:
v = u + at
We know the initial velocity (u) is 0 m/s, the final velocity (v) is 5.08 m/s, and the time (t1) is 1.5 s.
Substituting the values into the equation:
5.08 m/s = 0 m/s + a1 * 1.5 s
Rearranging the equation, we have:
5.08 m/s = 1.5 a1
Now, we have two equations:
3.4 m/s = 1.5 a1 (equation 1)
5.08 m/s = 1.5 a1 (equation 2)
We can solve these two equations simultaneously to find the value of a1.
Dividing equation 2 by equation 1:
5.08 m/s / 3.4 m/s = 1.5 a1 / 1.5 a1
Simplifying the equation, we get:
1.494 = a1 / a1
1.494 = 1
This means that the value of a1 is 1 m/s².
So, the value of the football player's acceleration during the initial 1.5-second period is 1 m/s².