Question

A train starting from rest picks up a speed of 10ms^-1 in 100s. it continues to move at the same speed for the next 250 sec. It is then brought to rest in the next 50 sec. Plot speed time graph for entire motion of the train. Calculate: Acceleration when the train is accelerating Retardation when it is retarding Total distance covered by the train.

Answer 1

➭ Questíon :-

$\longrightarrow$A train starting from rest picks up a speed of 10m/s in 100s. it continues to move at the same speed for the next 250 sec. It is then brought to rest in the next 50 sec. Plot speed time graph for entire motion of the train. Calculate: Acceleration when the train is accelerating Retardation when it is retarding Total distance covered by the train.

➭ Given :-

$\longrightarrow$Speed v = 10 m/s

$\longrightarrow$Time t = 100 s

➭ To find :-

$\longrightarrow$The acceleration of the train

$\longrightarrow$the retardation of the train

$\longrightarrow$the total distance covered by the train

➭ Solution :-

According to graph,

➛ The acceleration of the train is :

$\longrightarrow \rm Accerlation = \frac{change \: in \: velocity }{time} \\ \\ \longrightarrow \rm a = \frac{10 - 0}{100} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \boxed {\rm a = 0.1m {s}^{ - 2}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

➛ The retardation of the train is :

$\longrightarrow \rm Retardation = \frac{change \: in \: velocity}{time} \\ \\ \longrightarrow \rm a = \frac{0 - 10}{50} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \boxed {\rm a = - 0.2m {s}^{ - 2}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

➛ The total distance covered by the train is :

According to speed - time graph

Area under the curve given total distance

→ Total distance = Area of ∆OAB + area of ABCD + Area of ∆DCE

$\longrightarrow \rm distance = \frac{1}{2} \times 100 \times 10 + 2500 + \frac{1}{2} \times 50 \times 10 \\ \\ \longrightarrow \boxed {\rm distance = 3250m }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hence, The acceleration is 0.1 m/s² , the retardation is −0.2m/s² and the total distance is 3250 m.