Home›Irodov›Problem 1.4
Irodov 1.4Trickykinematicsaverage-velocitys-t-graphinstantaneous-velocity

Distance-Time Graph Analysis

Problem 1.4
A point moves rectilinearly in one direction. The distance ss traversed by the point as a function of time tt is shown in the graph. The graph has three phases: a concave-up curve (acceleration) from t=0t = 0 to t=10 st = 10 \text{ s}, a straight line (constant velocity) from t=10 st = 10 \text{ s} to t=14 st = 14 \text{ s}, and a concave-down curve (deceleration) from t=14 st = 14 \text{ s} to t=20 st = 20 \text{ s}, reaching s=200 cms = 200 \text{ cm}. Using the graph, find: (a) the average velocity over the whole time of motion; (b) the maximum velocity; (c) the time t0t_0 at which the instantaneous velocity equals the mean velocity averaged over the first t0t_0 seconds.
Answer: ⟨v⟩=10 cm/s,  vmax⁡=25 cm/s,  t0≈16 s\langle v \rangle = 10 \text{ cm/s},\; v_{\max} = 25 \text{ cm/s},\; t_0 \approx 16 \text{ s}
Solution Path
Reading the s-t graph: average velocity = total slope (10 cm/s), maximum velocity = steepest slope in the linear phase (25 cm/s), and the instantaneous-equals-mean condition is met when the tangent passes through the origin (t0 ~ 16 s).
01Problem Restatement
1/6
A point moves in a straight line in one direction. We are given the distance-time graph, which shows three distinct phases of motion: an accelerating phase (curved, concave up), a constant-velocity phase (straight line, steepest slope), and a decelerating phase (curved, concave down).Given:
- Total distance traveled: s=200 cms = 200 \text{ cm}
- Total time of motion: T=20 sT = 20 \text{ s}
- From t=0t = 0 to t=10 st = 10 \text{ s}: acceleration phase (concave up), covering 50 cm50 \text{ cm}
- From t=10 st = 10 \text{ s} to t=14 st = 14 \text{ s}: constant velocity phase (straight line), covering 100 cm100 \text{ cm}
- From t=14 st = 14 \text{ s} to t=20 st = 20 \text{ s}: deceleration phase (concave down), covering 50 cm50 \text{ cm}We need to find:
(a) The average velocity over the whole motion, in cm/s.
(b) The maximum velocity, in cm/s.
(c) The time t0t_0 at which the instantaneous velocity equals the mean velocity averaged over the first t0t_0 seconds.
Find ⟨v⟩\langle v \rangle, vmax⁡v_{\max}, and t0t_0 from the ss-tt graph
02Physical Picture and Strategy
2/6
Looking at the ss-tt graph, we see three phases of motion. In the first phase (00 to 10 s10 \text{ s}), the curve bends upward (concave up), meaning the slope (velocity) is increasing: the point is speeding up. In the second phase (1010 to 14 s14 \text{ s}), the graph is a straight line with the steepest slope: the point moves at constant (and maximum) velocity. In the third phase (1414 to 20 s20 \text{ s}), the curve bends downward (concave down), meaning the slope is decreasing: the point is slowing down.Key data points on the graph: (0,0)(0, 0), (10,50)(10, 50), (14,150)(14, 150), (20,200)(20, 200). During the constant-velocity phase, the point covers Δs=150−50=100 cm\Delta s = 150 - 50 = 100 \text{ cm} in Δt=14−10=4 s\Delta t = 14 - 10 = 4 \text{ s}.Strategy: We will read all three answers directly from the graph using the physical meaning of slopes on an ss-tt graph.
- Average velocity = slope of the straight line connecting the start and end points.
- Maximum velocity = steepest slope on the graph = the slope of the linear (constant velocity) segment.
- For part (c), we need the time when the tangent line at a point on the curve passes through the origin.
Three phases: accel (00-10 s10 \text{ s}), uniform (1010-14 s14 \text{ s}), decel (1414-20 s20 \text{ s})
03Step 1: Average Velocity (Part a)
3/6
The average velocity over the entire trip is the total distance divided by the total time. On the ss-tt graph, this equals the slope of the straight line drawn from the origin (0,0)(0, 0) to the final point (20 s,200 cm)(20 \text{ s}, 200 \text{ cm}).Using the definition of average velocity:
⟨v⟩=total distancetotal time\langle v \rangle = \frac{\text{total distance}}{\text{total time}}Substituting the values read from the graph:
⟨v⟩=200 cm20 s=10 cm/s\langle v \rangle = \frac{200 \text{ cm}}{20 \text{ s}} = 10 \text{ cm/s}This means: on average, the point covers 10 centimeters every second over the entire 20-second trip.
⟨v⟩=10 cm/s\langle v \rangle = 10 \text{ cm/s}
04Step 2: Maximum Velocity (Part b)KEY INSIGHT
4/6
The instantaneous velocity at any time equals the slope of the ss-tt curve at that point. The maximum velocity occurs where the slope is steepest.Looking at the graph, the steepest part is the straight-line segment from t=10 st = 10 \text{ s} to t=14 st = 14 \text{ s}. During this phase the velocity is constant (the slope does not change), and this constant velocity is the maximum velocity reached during the entire motion.During the acceleration phase the slope is smaller (the curve is still bending upward). During the deceleration phase the slope is decreasing. So the straight segment has the largest slope.Calculating the slope of the linear segment:
vmax⁡=ΔsΔt=s(14)−s(10)14−10v_{\max} = \frac{\Delta s}{\Delta t} = \frac{s(14) - s(10)}{14 - 10}vmax⁡=150 cm−50 cm4 s=100 cm4 s=25 cm/sv_{\max} = \frac{150 \text{ cm} - 50 \text{ cm}}{4 \text{ s}} = \frac{100 \text{ cm}}{4 \text{ s}} = 25 \text{ cm/s}This means: the point reaches a top speed of 25 cm/s during the constant-velocity phase, which is 2.5 times the average velocity.
vmax⁡=25 cm/sv_{\max} = 25 \text{ cm/s}
05Step 3: When Instantaneous Equals Mean (Part c)
5/6
We need to find the time t0t_0 at which the instantaneous velocity equals the mean velocity averaged over [0,t0][0, t_0].The mean velocity over [0,t0][0, t_0] is:
⟨v⟩0→t0=s(t0)t0\langle v \rangle_{0 \to t_0} = \frac{s(t_0)}{t_0}On the ss-tt graph, this is the slope of the straight line drawn from the origin OO to the point P=(t0,s(t0))P = (t_0, s(t_0)) on the curve.The instantaneous velocity at t0t_0 is:
v(t0)=dsdt∣t0v(t_0) = \frac{ds}{dt}\bigg|_{t_0}On the graph, this is the slope of the tangent line to the curve at point PP.The condition v(t0)=⟨v⟩0→t0v(t_0) = \langle v \rangle_{0 \to t_0} means the slope of the tangent at PP equals the slope of the line OPOP. Geometrically, this happens when the tangent line at PP passes through the origin OO.Sliding a point along the curve and checking this condition graphically, we find that the tangent passes through the origin at approximately t0≈16 st_0 \approx 16 \text{ s}.This means: at about 16 seconds into the motion, the point's current speed exactly equals its average speed up to that moment. Before this time the instantaneous speed was higher than the running average (during the fast phase), and after this time it drops below.
t0≈16 st_0 \approx 16 \text{ s}
06Verification and Summary
6/6
Verification: We check each answer against the graph.Part (a): ⟨v⟩=200 cm20 s=10 cm/s\langle v \rangle = \frac{200 \text{ cm}}{20 \text{ s}} = 10 \text{ cm/s} ✓\checkmarkPart (b): vmax⁡=100 cm4 s=25 cm/sv_{\max} = \frac{100 \text{ cm}}{4 \text{ s}} = 25 \text{ cm/s} ✓\checkmarkSanity check: vmax⁡>⟨v⟩v_{\max} > \langle v \rangle (25 > 10). This makes sense because the point is not always moving at top speed.Part (c): At t0≈16 st_0 \approx 16 \text{ s}, the tangent to the ss-tt curve passes through the origin. This is in the deceleration phase (1414-20 s20 \text{ s}), which is physically reasonable: the point has already passed through its fastest phase, and as it slows down, its instantaneous speed eventually drops to match the running average. ✓\checkmarkSummary: In this problem the key was reading physical quantities directly from the ss-tt graph. Average velocity is the slope from start to end. Maximum velocity is the steepest slope on the curve. The condition for instantaneous velocity equaling mean velocity translates to a beautiful geometric condition: the tangent at that point passes through the origin. The answers are 10 cm/s, 25 cm/s, and approximately 16 s.
⟨v⟩=10 cm/s,vmax⁡=25 cm/s,t0≈16 s\boxed{\langle v \rangle = 10 \text{ cm/s}, \quad v_{\max} = 25 \text{ cm/s}, \quad t_0 \approx 16 \text{ s}}
Want more practice?
Try more PYQs from this chapter →