Question

a train accelerates uniformly from 72km h^-1 to 90 km h^-1 .Find the distance travelled in 2 seconds

Please solve this with full explanation ​

Answer 1

$\bigstar$ EXPLANATION $\bigstar$

• Given

u (Initial speed of the train) = 72 km/hr = 72 * 5/18 = 20 m/s

v (Final speed of the train) = 90 km/hr = 90 * 5/18 = 25 m/s

t  = 2s

• Procedure

We know that $a = \frac{(v-u)}{t}$

Let's find the acceleration of the train now

$a = \frac{(25 - 20)}{2}\: m/s^2$

$a = \frac{5}{2}\:m/s^2$

$a = 2.5\:m/s^2$

Using the second equation of motion,

$s = ut + \frac{at^2}{2}$

$s = 20*2 + \frac{2.5*4}{2}$

$s = 40 + 5 = 45\:m$

• Concepts used

i) How to find acceleration

ii) Using Equation of motion

• Extra Information

i) $v = u+ at$

ii) $s = ut + \frac{at^2}{2}$

iii) $v^2 - u^2 = 2as$