A child takes reading of two cars running on highway, for his school

1.	A child takes reading of two cars running on highway, for his school project. He draws a position-time graph of the two cars as shown in the figure(i) What is the velocity of two cars when they meet together?(ii) What is the difference in velocities of the two cars when they cover their maximum distance?(iii) What will be acceleration of the two cars in first 20 s?
Answer» (i) According to graph, the velocity of two cars when they meet each other are, x = 70m t = 308 v = \(\frac{x}{t}=\frac{70}{30}\) = 2.33 m/s (ii) According to graph, for maximum distance. For 1^stcar, x₁ = 120m t₁ = 50 s v₁ = \(\frac{x_1}{t_1}=\frac{120}{50 }\) v₁ = 2.4 m/s For 2^nd car, x₂ = 90 m t₂ = 60 s v₂ = \(\frac{x_2}{t_2}=\frac{90}{60 }\) v₂ = 1.5 m/s Difference in velocities is given by, v₁ – v₂ = 2.4 – 1.5 = 0.9 m/s (iii) According to graph, Acceleration of 1^stcar in first 20 s v₁ = \(\frac{X_1}{t}\) v₁ = \(\frac{60}{20}\) v₁ = 3 m/s a₁ = \(\frac{V_1}{t}=\frac{3}{20}\) a₁ = 0.15 m/s² Acceleration of 2^nd car in first 20 s v₂ = \(\frac{X_2}{t}\) v₂ = \(\frac{40}{20}\) v₂ = 2 m/s a₂ = \(\frac{V_2}{t}=\frac{2}{20}\) a₂ = 0.1 m/s² Now, a₁ – a₂= 0.15 – 0.1 = 0.05 m/s² (i) The velocity of two cars when they meet together is 2.33 m/s. (ii) The difference in velocities of two cars when they cover maximum distance is 0.9 m/s. (iii) The accelerator of two cars in 20 s is 0.05 m/s².

A child takes reading of two cars running on highway, for his school project. He draws a position-time graph of the two cars as shown in the figure(i) What is the velocity of two cars when they meet together?(ii) What is the difference in velocities of the two cars when they cover their maximum distance?(iii) What will be acceleration of the two cars in first 20 s?

Answer»

(i) According to graph, the velocity of two cars when they meet each other are,

x = 70m

t = 308

v = \(\frac{x}{t}=\frac{70}{30}\) = 2.33 m/s

(ii) According to graph, for maximum distance.

For 1^stcar,

x₁ = 120m

t₁ = 50 s

v₁ = \(\frac{x_1}{t_1}=\frac{120}{50 }\)

v₁ = 2.4 m/s

For 2^nd car,

x₂ = 90 m

t₂ = 60 s

v₂ = \(\frac{x_2}{t_2}=\frac{90}{60 }\)

v₂ = 1.5 m/s

Difference in velocities is given by,

v₁ – v₂ = 2.4 – 1.5 = 0.9 m/s

(iii) According to graph,

Acceleration of 1^stcar in first 20 s

v₁ = \(\frac{X_1}{t}\)

v₁ = \(\frac{60}{20}\)

v₁ = 3 m/s

a₁ = \(\frac{V_1}{t}=\frac{3}{20}\)

a₁ = 0.15 m/s²

Acceleration of 2^nd car in first 20 s

v₂ = \(\frac{X_2}{t}\)

v₂ = \(\frac{40}{20}\)

v₂ = 2 m/s

a₂ = \(\frac{V_2}{t}=\frac{2}{20}\)

a₂ = 0.1 m/s²

Now,

a₁ – a₂= 0.15 – 0.1

= 0.05 m/s²

(i) The velocity of two cars when they meet together is 2.33 m/s.

(ii) The difference in velocities of two cars when they cover maximum distance is 0.9 m/s.

(iii) The accelerator of two cars in 20 s is 0.05 m/s².