Question

Q7. A train accelerates from 36 3 points
km/h to 54 km/h in 10 sec.(i)
Acceleration(ii) The distance
travelled by car.Answer
O
a = 5 m/s2, s = 100 m
O a = -5 m/s2, s = 125 m
a = 15 m/s2, s = 120 m
O
a = 5 m/s2, s = 125 m​

Answer 1

Explanation:

Given

• initial velocity of the train,u = 36 kmph

we should convert into m/s

we know that 1.kmph = 5/18 m/s

so,36 kmph = 36 × 5/18 m/s

= 10 m/s

• final velocity of the train,v = 54 kmph

we should convert into m/s

we know that 1 kmph = 5/18 m/s

54 kmph = 54 × 5/18 m/s

54 kmph = 15 m/s

• time,t = 10s

To find

• acceleration of the car
• distance travelled by the car

Equations used

• v = u + at
• $\sf{v}^{2} - {u}^{2}$= 2as

Substitute the values of v,u,t in equation of motion i.e v = u+at

15 = 10 + 10a

10a = 5

a = 5/10

a = 0.5 m/s^2

Now substitute values of v,u,a in $\sf{v}^{2} - {u}^{2}$ = 2as

15^2 - 10^2 = 2(0.5)(s)

225 - 100 = s

s = 125m

HOPE IT HELPS...