A car accelerates from rest at a steady rate of 1 ms-2. Calculate:
(a) the time taken to reach 15 ms-1
(b) the distance travelled during this time
(c) the velocity of the car when it was 100 m from the start point
a. V = at = 15 m/s.
1*t = 15,
t = 15 s.
b. d = 0.5a*t^2,
d = 0.5*(15)^2 = 112.5 m.
c. V^2 = Vo^2 + 2a*d,
V^2 = 0 + 2*100 = 200,
V = 14.14 m/s.
To solve these problems, we can use the equations of motion for uniformly accelerated motion.
(a) Time taken to reach 15 m/s:
We can use the equation:
v = u + at
where
v = final velocity = 15 m/s
u = initial velocity = 0 m/s (as the car starts from rest)
a = acceleration = 1 m/s²
t = time taken (unknown)
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values, we find:
t = (15 m/s - 0 m/s) / (1 m/s²)
t = 15 s
Therefore, the time taken to reach 15 m/s is 15 seconds.
(b) Distance traveled during this time:
We can use the equation:
s = ut + (1/2)at²
where
s = distance traveled (unknown)
u = initial velocity = 0 m/s
t = time taken = 15 s
a = acceleration = 1 m/s²
Substituting the given values, we find:
s = (0 m/s) * (15 s) + (1/2) * (1 m/s²) * (15 s)²
s = 0 + 0.5 * 1 * 225
s = 0 + 112.5
s = 112.5 m
Therefore, the distance traveled during this time is 112.5 meters.
(c) Velocity of the car when it was 100 meters from the start point:
We can use the equation:
v² = u² + 2as
Rearranging the equation, we have:
v² = u² + 2as
v² = (0 m/s)² + 2 * (1 m/s²) * (100 m)
v² = 0 + 2 * 1 * 100
v² = 200
v = √200
v ≈ 14.14 m/s
Therefore, the velocity of the car when it was 100 meters from the start point is approximately 14.14 m/s.
To solve these problems, we can use the equations of motion that relate acceleration, time, initial velocity, final velocity, and distance.
(a) To find the time taken to reach a velocity of 15 m/s, we can use the equation:
v = u + at
where:
v = final velocity = 15 m/s
u = initial velocity = 0 m/s (as the car starts from rest)
a = acceleration = 1 m/s^2
t = time taken (unknown)
Rearranging the equation, we have:
t = (v - u) / a
Plugging in the values, we get:
t = (15 - 0) / 1 = 15 s
Therefore, the car takes 15 seconds to reach a velocity of 15 m/s.
(b) To find the distance traveled during this time, we can use the equation:
s = ut + 0.5at^2
where:
s = distance traveled (unknown)
u = initial velocity = 0 m/s
a = acceleration = 1 m/s^2
t = time taken = 15 s
Plugging in the values, we have:
s = (0 x 15) + 0.5 x 1 x (15^2) = 0 + 112.5 = 112.5 m
Therefore, the car travels a distance of 112.5 meters during this time.
(c) To find the velocity of the car when it's 100 m from the start point, we can use the equation:
s = ut + 0.5at^2
where:
s = distance traveled = 100 m (unknown)
u = initial velocity = 0 m/s
a = acceleration = 1 m/s^2
t = time taken (unknown)
Rearranging the equation, we have a quadratic equation:
0.5at^2 + ut - s = 0
Plugging in the values, we get:
0.5 x 1 x t^2 + 0 x t - 100 = 0
Solving this quadratic equation, we find two values for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case:
a = 0.5
b = 0
c = -100
Plugging in the values, we get:
t = (0 ± √(0^2 - 4 x 0.5 x -100)) / (2 x 0.5)
t = (0 ± √(0 + 200)) / 1
t = (0 ± √200) / 1
t = ± √200
Since time cannot be negative in this context, we take the positive value:
t = √200
Therefore, the time taken to reach a distance of 100 m is approximately √200 seconds.
To find the velocity at this point, we can substitute the value of t into the equation:
v = u + at
Plugging in the values, we have:
v = 0 + 1 x √200 = √200 m/s
Therefore, the velocity of the car when it was 100 m from the starting point is approximately √200 m/s.