A car moving at 10m/s when it begins to be accelerated at 2.5m/s^2. (a)How long does the car take to reach a speed of 25m/s? (b) How far does it go during this period
To find out how long it takes for the car to reach a speed of 25m/s, we can use the formula:
v = u + at
where:
v = final velocity = 25m/s
u = initial velocity = 10m/s
a = acceleration = 2.5m/s^2
t = time taken
(a) Solving for t:
25 = 10 + (2.5)t
25 - 10 = 2.5t
15 = 2.5t
t = 15/2.5
t = 6 seconds
So, it takes 6 seconds for the car to reach a speed of 25m/s.
To find out how far the car goes during this period, we can use the formula:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity = 10m/s
t = time taken = 6 seconds
a = acceleration = 2.5m/s^2
(b) Solving for s:
s = (10)(6) + (1/2)(2.5)(6)^2
s = 60 + (1/2)(2.5)(36)
s = 60 + (1/2)(90)
s = 60 + 45
s = 105 meters
So, the car travels 105 meters during this period.
To answer this question, we can use the equations of motion to find the time taken and the distance traveled by the car.
(a) The equation to calculate the time taken is:
𝑣 = 𝑢 + 𝑎𝑡
Where:
𝑣 = final velocity (25m/s)
𝑢 = initial velocity (10m/s)
𝑎 = acceleration (2.5m/s^2)
𝑡 = time taken
Rearranging the equation, we get:
𝑡 = (𝑣 - 𝑢) / 𝑎
Substituting the known values, we find:
𝑡 = (25 - 10) / 2.5
𝑡 = 15 / 2.5
𝑡 = 6 seconds
Therefore, it takes the car 6 seconds to reach a speed of 25m/s.
(b) The equation to calculate the distance traveled is:
𝑠 = 𝑢𝑡 + 0.5𝑎𝑡^2
Where:
𝑠 = distance traveled
𝑢 = initial velocity (10m/s)
𝑡 = time taken (6 seconds)
𝑎 = acceleration (2.5m/s^2)
Substituting the known values, we find:
𝑠 = (10 * 6) + 0.5 * 2.5 * (6^2)
𝑠 = 60 + 0.5 * 2.5 * 36
𝑠 = 60 + 0.5 * 90
𝑠 = 60 + 45
𝑠 = 105 meters
Therefore, the car travels a distance of 105 meters during this period.
vf=vi+at solve for t
d=vi*t+1/2 a t^2