Uniform Acceleration Motion: Solutions to Problems
Detailed solutions to problems on velocity and uniform acceleration are presented. These problems demonstrate applications of kinematic equations in physics. Tutorials on acceleration are also available.
Problem 1
From rest, a car accelerates at \(8 \, \text{m/s}^2\) for \(10 \, \text{s}\).
a) What is the car's position after \(10 \, \text{s}\)?
b) What is the car's velocity after \(10 \, \text{s}\)?
Solution
a) The car starts from rest: initial velocity \(u = 0\). Assuming initial position \(x_0 = 0\), position is given by:
\[x = \frac{1}{2} a t^2\]
where \(a = 8 \, \text{m/s}^2\), \(t = 10 \, \text{s}\):
\[x = \frac{1}{2} (8) (10)^2 = 400 \, \text{m}\]
b) Velocity after \(10 \, \text{s}\):
\[v = u + at = 0 + (8)(10) = 80 \, \text{m/s}\]
Problem 2
With initial velocity \(20 \, \text{km/h}\), a car accelerates at \(8 \, \text{m/s}^2\) for \(10 \, \text{s}\).
a) Position after \(10 \, \text{s}\)?
b) Velocity after \(10 \, \text{s}\)?
Solution
a) Initial velocity conversion:
\[u = 20 \, \frac{\text{km}}{\text{h}} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} \approx 5.56 \, \text{m/s}\]
Position equation:
\[x = ut + \frac{1}{2}at^2\]
\[x = (5.56)(10) + \frac{1}{2}(8)(10)^2 = 55.6 + 400 = 455.6 \, \text{m}\]
b) Final velocity:
\[v = u + at = 5.56 + (8)(10) = 85.56 \, \text{m/s}\]
Problem 3
A car accelerates uniformly from \(0\) to \(72 \, \text{km/h}\) in \(11.5 \, \text{s}\).
a) Acceleration in \(\text{m/s}^2\)?
b) Distance traveled during acceleration?
Solution
a) Convert final velocity:
\[v = 72 \, \frac{\text{km}}{\text{h}} \times \frac{1000}{3600} = 20 \, \text{m/s}\]
\[a = \frac{v - u}{t} = \frac{20 - 0}{11.5} \approx 1.74 \, \text{m/s}^2\]
b) Using average velocity:
\[x = \frac{1}{2}(u + v)t = \frac{1}{2}(0 + 20)(11.5) = 115 \, \text{m}\]
Or using position equation:
\[x = \frac{1}{2}at^2 = \frac{1}{2}(1.74)(11.5)^2 \approx 115 \, \text{m}\]
Problem 4
An object is thrown downward from a building at \(20 \, \text{m/s}\), hitting ground at \(40 \, \text{m/s}\).
a) Building height?
b) Time in air?
Solution
Take upward as positive, so \(u = -20 \, \text{m/s}\), \(v = -40 \, \text{m/s}\), \(a = -g = -9.8 \, \text{m/s}^2\).
a) Using:
\[v^2 = u^2 + 2a(x - x_0)\]
\[(-40)^2 = (-20)^2 + 2(-9.8)(0 - x_0)\]
\[1600 = 400 + 19.6x_0\]
\[x_0 = \frac{1200}{19.6} \approx 61.2 \, \text{m}\]
b) Using:
\[x - x_0 = \frac{1}{2}(u + v)t\]
\[-61.2 = \frac{1}{2}(-20 - 40)t\]
\[t = \frac{61.2}{30} = 2.04 \, \text{s}\]
Problem 5
A train decelerates from \(40 \, \text{m/s}\) to stop over \(100 \, \text{m}\).
a) Acceleration?
b) Stopping time?
Solution
a) Using:
\[v^2 = u^2 + 2ax\]
\[0^2 = 40^2 + 2a(100)\]
\[a = -\frac{1600}{200} = -8 \, \text{m/s}^2\]
b) Method 1:
\[x = \frac{1}{2}(u + v)t \Rightarrow 100 = \frac{1}{2}(40 + 0)t \Rightarrow t = 5 \, \text{s}\]
Method 2:
\[x = ut + \frac{1}{2}at^2 \Rightarrow 100 = 40t - 4t^2\]
\[4t^2 - 40t + 100 = 0 \Rightarrow t^2 - 10t + 25 = 0 \Rightarrow (t-5)^2 = 0 \Rightarrow t = 5 \, \text{s}\]
Problem 6
A bicycle increases velocity from \(5 \, \text{m/s}\) to \(20 \, \text{m/s}\) in \(10 \, \text{s}\).
a) Acceleration?
b) Distance covered?
Solution
a) Acceleration:
\[a = \frac{v - u}{t} = \frac{8 - 5}{10} = 0.3 \, \text{m/s}^2\]
b) Distance:
\[x = \frac{1}{2}(u + v)t = \frac{1}{2}(5 + 8)(10) = 65 \, \text{m}\]
Or:
\[x = ut + \frac{1}{2}at^2 = 5(10) + \frac{1}{2}(0.3)(10)^2 = 50 + 75 = 65 \, \text{m}\]
Problem 7
a) Time for airplane to reach \(350 \, \text{km/h}\) over \(600 \, \text{m}\) from rest?
b) Acceleration during takeoff?
Solution
a) Convert final velocity:
\[v = 350 \, \frac{\text{km}}{\text{h}} = \frac{350 \times 1000}{3600} \approx 97.2 \, \text{m/s}\]
\[x = \frac{1}{2}(u + v)t \Rightarrow 600 = \frac{1}{2}(0 + 97.2)t\]
\[t = \frac{600}{48.6} \approx 12.3 \, \text{s}\]
b) Acceleration:
\[a = \frac{v}{t} = \frac{97.2}{12.3} \approx 7.9 \, \text{m/s}^2\]
Problem 8
Object starts \(20 \, \text{m}\) left of origin at \(10 \, \text{m/s}\), accelerates right at \(4 \, \text{m/s}^2\) for \(5 \, \text{s}\). Find final position.
Solution
\[x = x_0 + ut + \frac{1}{2}at^2 = -20 + 10(5) + \frac{1}{2}(4)(5)^2\]
\[x = -20 + 50 + 50 = 80 \, \text{m} \quad \text{(right of origin)}\]
Problem 9
Minimum runway length for airplane landing at \(360 \, \text{km/h}\) with deceleration \(-10 \, \text{m/s}^2\)?
Solution
\[u = 360 \, \frac{\text{km}}{\text{h}} = \frac{360 \times 1000}{3600} = 100 \, \text{m/s}\]
\[v^2 = u^2 + 2ax \Rightarrow 0 = 100^2 + 2(-10)x\]
\[x = \frac{10000}{20} = 500 \, \text{m}\]
Problem 10
Rock dropped into well; splash heard \(8 \, \text{s}\) later. Speed of sound = \(340 \, \text{m/s}\). Well depth?
Solution
Let \(t\) = fall time, \(8-t\) = sound return time. Depth \(h\):
\[h = \frac{1}{2}gt^2 = 4.9t^2 \quad \text{(falling)}\]
\[h = 340(8 - t) \quad \text{(sound)}\]
\[4.9t^2 = 2720 - 340t\]
\[4.9t^2 + 340t - 2720 = 0\]
Solving: \(t \approx 7.24 \, \text{s}\) (positive root).
\[h = 340(8 - 7.24) \approx 258 \, \text{m}\]
Problem 11
Rock thrown upward reaches \(10 \, \text{m}\) height.
a) Time in air?
b) Initial velocity?
Solution
At highest point \(v=0\), \(a=-g=-9.8 \, \text{m/s}^2\).
a) Using:
\[x = \frac{1}{2}(u + v)t \Rightarrow 10 = \frac{1}{2}(u + 0)t\]
and \(v = u + at \Rightarrow 0 = u - 9.8t \Rightarrow u = 9.8t\)
\[10 = \frac{1}{2}(9.8t)t = 4.9t^2 \Rightarrow t = \sqrt{\frac{10}{4.9}} \approx 1.43 \, \text{s}\]
b) Initial velocity:
\[u = 9.8t \approx 9.8 \times 1.43 \approx 14.0 \, \text{m/s}\]
Problem 12
A car accelerates from rest at \(1.0 \, \text{m/s}^2\) for \(20.0 \, \text{s}\) , moves at constant speed for half an hour, then decelerates to stop in \(30.0 \, \text{s}\). Find the total distance covered by the car.
Solution
Stage 1: Acceleration
\[x_1 = \frac{1}{2}at^2 = \frac{1}{2}(1)(20)^2 = 200 \, \text{m}\]
\[v = at = 1 \times 20 = 20 \, \text{m/s}\]
Stage 2: Constant velocity
\[t_2 = 0.5 \, \text{h} = 1800 \, \text{s}\]
\[x_2 = vt_2 = 20 \times 1800 = 36,000 \, \text{m}\]
Stage 3: Deceleration
\[x_3 = \frac{1}{2}(u + v)t = \frac{1}{2}(20 + 0)(30) = 300 \, \text{m}\]
Total distance:
\[x_{\text{total}} = 200 + 36,000 + 300 = 36,500 \, \text{m}\]
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