Question

calculate acceleration for a body increasing its velocity from 56 kilometre per hour to 90 km per hour in 20 seconds what is the distance covered by the body​

Answer 1

Initial velocity = 56 km/hr

Final velocity = 90 km/hr

Time taken = 20 secs

Since the acceleration is assumed to be constant during the time of observation , we can apply equation of kinematics.

We need to convert the unit of speed from km/hr to m/s :

56 km/hr = 56 × (5/18) = 15.55 m/s

90 km/hr = 90 × (5/18) = 25 m/s

Now ,

$v = u + at$

$= > 25= 15.55+ (a \times 20)$

$= > 20a = 9.45$

$= > a = 0.4775 \: m {s}^{ - 2}$

Now calculation of distance travelled :

${v}^{2} = {u}^{2} + 2as$

$= > {(25)}^{2} = {(15.55)}^{2} + 2 \times (0.4775) \times s$

$= > 383.1975 = 0.955s$

$= > s = 401.25 \: m$

Answer 2

▶ Solution :

Given :

▪ Initial velocity of body = 56kmph

▪ Final velocity of body = 90kmph

▪ Time interval = 20s

To Find :

▪ Distance covered by body in the given time interval.

Concept :

▪ First, we have to find out acceleration of body.

▪ Since acceleration has constant magnitude throughout the whole journey, we can directly calculate the distance by second equation of kinematics.

▪ Acceleration is defined as ratio of change in velocity to the time interval.

▪ Acceleration is a vector quantity.

Conversion :

✏ 1kmph = 5/18mps

✏ 56kmph = 56×5/18 = 15.56mps

✏ 90kmph = 90×5/18 = 25mps

Calculation :

$\leadsto\bf\:a=\dfrac{v-u}{t}\\ \\ \leadsto\sf\:a=\dfrac{25-15.56}{20}\\ \\ \leadsto\bf\red{a=0.47\:ms^{-2}}\\ \\ \leadsto\bf\:v^2=u^2+2as\\ \\ \leadsto\sf\:(25)^2=(15.56)^2+2(0.47)s\\ \\ \leadsto\sf\:s=\dfrac{625-242.11}{0.94}\\ \\ \leadsto\boxed{\bf{\orange{s=407.32\:m}}}$