Question

The driver of a TGV travelling at a speed of 90m/s sights a truck on the railway track at a distance of 1 km ahead. Then he applies the brakes to decelerate the train at a rate of 5m/s^2. What is the distance travelled by the train before coming to rest? Will the train collide with the truck? ​

Answer 1

Given:

Initial velocity of TGV, u= 90 m/s

Final velocity of TGV, v= 0 m/s

⇪ (Since it stopped after applying brake)

Acceleration of car, a= -5 ms⁻²

⇪ (Negative sign shows that it is deaccelerating)

Distance of Truck from the point driver sees the truck, d= 1 km= 1000m    

To Find:

a) Distance travelled by train before coming to rest.

b) Will the train collide with the truck

Solution:

Let the distance covered by TGV after applying brake be 's' m

We know that,

• According to third equation of motion for constant acceleration

$\pink{\boxed{\bf{v^{2}-u^{2}=2as}}}$

where,

v is the final velocity of body

u is the initial velocity of body

a is acceleration of body

s is displacement of body

Now, on applying third equation of motion on given car, we get

$\longrightarrow\mathrm{v^{2}-u^{2}=2as}$

$\longrightarrow\mathrm{(0)^{2}-(90)^{2}=2(-5)s}$

$\longrightarrow\mathrm{-8100=-10s}$

$\longrightarrow\mathrm{-10s=-8100}$

$\longrightarrow\mathrm{s=\dfrac{-8100}{-10}}$

$\longrightarrow\mathrm{\green{s=810\:m}}$

Now,

Distance from the truck before which the TGV stops

= d - s

= 1000 - 810

= 190 m

Therefore, TGV will stops at a distance of 190 m before the Truck. Thus, TGV will not collide with Truck.

Hence, the TGV has travelled 810 m before coming to rest and has not collide with truck.

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Answer 2

Explanation:

Initial velocity of TGV, u= 90 m/s

Final velocity of TGV, v= 0 m/s

⇪ (Since it stopped after applying brake)

Acceleration of car, a= -5 ms⁻²

⇪ (Negative sign shows that it is deaccelerating)

Distance of Truck from the point driver sees the truck, d= 1 km= 1000m

To Find:

a) Distance travelled by train before coming to rest.

b) Will the train collide with the truck

Solution:

Let the distance covered by TGV after applying brake be 's' m

We know that,

According to third equation of motion for constant acceleration

\pink{\boxed{\bf{v^{2}-u^{2}=2as}}}

v

2

−u

2

=2as

where,

v is the final velocity of body

u is the initial velocity of body

a is acceleration of body

s is displacement of body

Now, on applying third equation of motion on given car, we get

\longrightarrow\mathrm{v^{2}-u^{2}=2as}⟶v

2

−u

2

=2as

\longrightarrow\mathrm{(0)^{2}-(90)^{2}=2(-5)s}⟶(0)

2

−(90)

2

=2(−5)s

\longrightarrow\mathrm{-8100=-10s}⟶−8100=−10s

\longrightarrow\mathrm{-10s=-8100}⟶−10s=−8100

\longrightarrow\mathrm{s=\dfrac{-8100}{-10}}⟶s=

−10

−8100

\longrightarrow\mathrm{\green{s=810\:m}}⟶s=810m

Now,

Distance from the truck before which the TGV stops

= d - s

= 1000 - 810

= 190 m

Therefore, TGV will stops at a distance of 190 m before the Truck. Thus, TGV will not collide with Truck.

Hence, the TGV has travelled 810 m before coming to rest and has not collide with truck.