Question

Figure 3.E2 shows the velocity-time graph for a
particle moving in a fixed direction. (a) Find the
acceleration of the particle. (b) If the mass of the
particle is 200 g, what is the force acting on it?
Velocity (m/s)
0
1
2 3 Time (seconds)
Fig. 3.E2​

Answer 1

Answer:

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a=

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= Δt

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv =

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv = 5

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv = 520

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv = 520

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv = 520 =4m/s

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv = 520 =4m/s 2

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv = 520 =4m/s 2

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv = 520 =4m/s 2 Mass of the particle, m=100g=0.1kg

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv = 520 =4m/s 2 Mass of the particle, m=100g=0.1kgHence, force acting on the particle, F=ma=0.1kg×4m/s

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv = 520 =4m/s 2 Mass of the particle, m=100g=0.1kgHence, force acting on the particle, F=ma=0.1kg×4m/s 2

From the graph, we can say that for time interval Δt=5s, change in velocity, Δv=20m/s.Therefore, acceleration, a= ΔtΔv = 520 =4m/s 2 Mass of the particle, m=100g=0.1kgHence, force acting on the particle, F=ma=0.1kg×4m/s 2 =0.4N