A train is running at full speed when brakes are applied. In the first minute, it travels 8 km, and in the next minute, it travels 3 km. Assuming constant retardation, the initial speed of the train is:
A. 150 m/s
B. 175 m/s
C. 200 m/s
D. 225 m/s
Answer
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Hint: The initial velocity is the velocity with which we start the observation on a body. In this case, the train is moving at some speed. We have to find this initial speed. When the brakes are applied on the train, it starts slowing down, and eventually, it stops after a certain distance, called stopping distance.
Formula used: $s= ut+\dfrac12 a t^2$
Complete step-by-step solution:
Here we are asked about the distance, giving information about the time. Hence we can use any equation of motion to calculate the parameters but the easiest would be using the second equation of motion i.e. $s= ut+\dfrac12 a t^2$.
Now, let the initial velocity of the train be ‘u’ and the retardation is ‘a’.
In first case:
t=1 min = 60 sec
and s= 8000 m:
Hence applying $s= ut+\dfrac12 a t^2$
$8000= u(60)+\dfrac12 a (60)^2$
$a=\dfrac{16000 - (120)u}{60^2}$
Also, in second case:
t=1+1 = 2 min = 120 sec
s = 8000 + 3000 = 11000 m:
Thus, applying:$s= ut+\dfrac12 a t^2$
$11000 = u(120)+\dfrac12 a (120)^2$
Now, putting the value of ‘a’ in this equation:
$\Rightarrow 11000 = u(120)+\dfrac12 (\dfrac{16000 - (120)u}{60^2}) (120)^2$
$\Rightarrow 11000 = u(120)+(32000-(240)u)$
$\Rightarrow 11000 = -120u+32000$
$\Rightarrow u= \dfrac{32000-11000}{120} = 175 m/s$
Hence, option B. is correct.
Additional information:
Retardation is the kind acceleration, which opposes the motion of the body. This is also called negative acceleration. This is caused in the case of vehicles when the brakes are applied. When the brakes are applied, the friction between the road and tire tends to move at the same speed this causes the skidding of the tire and this opposes the motion of the vehicle.
Note: It must be noted that for the second case when the distance traveled is 3 km, we can apply this data also in the equation of motion. Just the difference is that that time, we’ll take the initial velocity of the train, not ‘u’ but ‘u+at’, where ‘a’ is the negative acceleration, since due to retardation, the speed goes on decreasing continuously.
Formula used: $s= ut+\dfrac12 a t^2$
Complete step-by-step solution:
Here we are asked about the distance, giving information about the time. Hence we can use any equation of motion to calculate the parameters but the easiest would be using the second equation of motion i.e. $s= ut+\dfrac12 a t^2$.
Now, let the initial velocity of the train be ‘u’ and the retardation is ‘a’.
In first case:
t=1 min = 60 sec
and s= 8000 m:
Hence applying $s= ut+\dfrac12 a t^2$
$8000= u(60)+\dfrac12 a (60)^2$
$a=\dfrac{16000 - (120)u}{60^2}$
Also, in second case:
t=1+1 = 2 min = 120 sec
s = 8000 + 3000 = 11000 m:
Thus, applying:$s= ut+\dfrac12 a t^2$
$11000 = u(120)+\dfrac12 a (120)^2$
Now, putting the value of ‘a’ in this equation:
$\Rightarrow 11000 = u(120)+\dfrac12 (\dfrac{16000 - (120)u}{60^2}) (120)^2$
$\Rightarrow 11000 = u(120)+(32000-(240)u)$
$\Rightarrow 11000 = -120u+32000$
$\Rightarrow u= \dfrac{32000-11000}{120} = 175 m/s$
Hence, option B. is correct.
Additional information:
Retardation is the kind acceleration, which opposes the motion of the body. This is also called negative acceleration. This is caused in the case of vehicles when the brakes are applied. When the brakes are applied, the friction between the road and tire tends to move at the same speed this causes the skidding of the tire and this opposes the motion of the vehicle.
Note: It must be noted that for the second case when the distance traveled is 3 km, we can apply this data also in the equation of motion. Just the difference is that that time, we’ll take the initial velocity of the train, not ‘u’ but ‘u+at’, where ‘a’ is the negative acceleration, since due to retardation, the speed goes on decreasing continuously.
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