If a car with a velocity of 2.0m/s at t=0 accelerates at a rate of 4.0m/s2 for 2.5 sec, what is its velocity
at a t=2.5s?
Using basic Calculus:
a = 4 m/s^2
v = 4t + c
when t = 0, v = 2 m/s
2 = 0 + c ---> c = 2
v = 4t + 2 <-------- valid for 0 ≤ t ≤ 2.5
when t = 2.5
v = 4(2.5) + 2 = 12
at t = 2.5 sec, v = 12 m/s
To find the velocity of the car at t=2.5s, we can use the equation for linear motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 2.0 m/s (initial velocity)
a = 4.0 m/s^2 (acceleration)
t = 2.5 s (time)
Substituting the values into the equation, we have:
v = 2.0 m/s + (4.0 m/s^2)(2.5 s)
First, we need to calculate the value inside the parentheses:
(4.0 m/s^2)(2.5 s) = 10.0 m/s
Now we substitute this value back into the equation:
v = 2.0 m/s + 10.0 m/s
Adding the velocities, we get:
v = 12.0 m/s
Therefore, the velocity of the car at t=2.5s is 12.0 m/s.
To find the velocity of the car at t=2.5s, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 2.0 m/s (initial velocity)
a = 4.0 m/s^2 (acceleration)
t = 2.5 s (time)
Substituting the given values into the equation, we have:
v = 2.0 m/s + 4.0 m/s^2 x 2.5 s
Calculating the equation, we get:
v = 2.0 m/s + 10 m/s
v = 12.0 m/s
Therefore, the velocity of the car at t=2.5s is 12.0 m/s.