Question

A train accelerates from 40 km/h to 548km/h in 10 sec.
(i) Acceleration
(ii) The distance travelled by car.

Answer 1

Answer:

125 m

Explanation:

Given:

Initial velocity= 36km/h=36x5/18=10m/s

Final velocity =54km/h=54x5/18=15m/s

Time =10sec

Acceleration = v-u/ t

=15-10/10=5/10=1/2=0.5 m/s2

Distance =s=?

From second equation of motion:

S=ut +1/2 at^2

=10*10+1/2*0.5*10*10

=100+25

=125m

So distance travelled is 125m

Answer 2

$\huge{\underline{\tt{\red{Find-}}}}$

• (i) Acceleration [a]
• (ii) The distance travelled by car [s]

$\huge\underline{\overline{\mid{\bold{\orange{Given-}}\mid}}}$

• Initial spped u= 40km/h = 11.1m/s
• Final speed v= 548km/h = 152.2m/s
• Time t= 10 sec

$\huge\underline{\overline{\mid{\bold{\blue{Solution-}}\mid}}}$

we know the formula of acceleration,

$a= \dfrac{v-u}{t} = \dfrac{152.2-11.1}{10}$

$\boxed{a= 14.11m/s^2}$

now, by using equation of motion we will find distance [s]

$s= ut + \dfrac{1}{2} at^2$

$s= 11.1\times 10 +\dfrac{1}{2}\times 14.11 \times 10\times 10$

$s= 111+705.5= 816.5$

$\boxed{s= 816.5m}$