Question

calculate the distance travelled by train which starts from rest and acquires a acc. of 5 m/s^2 in 13 s

Answer 1

Answer:

422.5m

Explanation:

s=ut+1/2 at²

s=0×13+ 1/2×5×13×13

s=1/2×5×169

s=422.5m

Answer 2

Information provided with us:

• Acceleration of 5 m/s^2 in 13 s by the train had been taken.

What we have to calculate:

• We have to calculate the distance travelled by train

Performing calculations:

Here the train has acquired a Acceleration (a) of 5ms-² and time taken is 13s from starting.

As we know that intial velocity (u) is always zero.

Thus, inorder to calculate the displacement ie., distance we would be using second equation of motion:

• $\red{\boxed{ \bf{S \: = \: ut + \dfrac{1}{2} at {}^{2} }}}$

Here in this equation,

• s is displacement
• u is initial velocity
• a is acceleration
• t is time taken

$\pink\bigstar \: \underline{\underline{\sf{Substituting \: the \: values:- }}}$

We already have,

• a = 5ms²
• t = 13s
• u = 0

Lets do now!

$: \longmapsto \: \sf{s \: = \: 0(13) + \dfrac{1}{2}(5)(13) {}^{2} }$

$: \longmapsto \: \sf{s \: = \: 0 \times 13+ \dfrac{1}{2}(5)(13) {}^{2} }$

$: \longmapsto \: \sf{s \: = \: 0 \times 13+ \dfrac{1}{2} \times 5(13) {}^{2} }$

$: \longmapsto \: \sf{s \: = \: 0 \times 13+ \dfrac{1}{2} \times 5 \times (13) {}^{2} }$

$: \longmapsto \: \sf{s \: = \: 0 \times 13+ \dfrac{1}{2} \times 5 \times13 \times 13 }$

$: \longmapsto \: \sf{s \: = \: \dfrac{1}{2} \times 5 \times13 \times 13 }$

$: \longmapsto \: \sf{s \: = \: \dfrac{1}{2} \times 5 \times 169}$

$: \longmapsto \: \sf{s \: = \: \dfrac{1}{2} \times 845}$

$: \longmapsto \: \sf{s \: = \: \dfrac{845 \times 1}{2}}$

$: \longmapsto \: \sf{s \: = \: \dfrac{845}{2}}$

• On cancelling it with 2,

$: \longmapsto \: \sf{s \: = \: \cancel\dfrac{845}{2}}$

$: \longmapsto \: \red{\boxed{\bf{s \: = \: 422.5}}}$

• Therefore, distance travelled by the train is of 422.5 m

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