Question

A car acceleratrs uniformly from
28 km h -¹ to 36 km h-¹ in 5s calculate ( i ) the acceleration and ( ii ) the distance covered by the car in that time

Answer 1

Given :

▪ Initial velocity = 28kmph

▪ Final velocity = 36kmph

▪ Time interval = 5s

To Find :

▪ Acceleration of car.

▪ Distance covered by car in the given interval of time.

Concept :

↗ Acceleration is defined as the rate of change in velocity.

↗ It is a vector quntity.

↗ It has both magnitude as well as direction.

↗ It can be positive, negative and zero.

↗ SI unit : m/s²

Conversion :

❇ 1kmph = 5/18mps

❇ 28kmph = 28×5/18 = 7.78mps

❇ 36kmph = 36×5/18 = 10mps

Calculation :

⚽ Acceleration :

$\dashrightarrow\sf\:a=\dfrac{v-u}{t}\\ \\ \dashrightarrow\sf\:a=\dfrac{10-7.78}{5}\\ \\ \dashrightarrow\sf\:a=\dfrac{2.22}{5}\\ \\ \dashrightarrow\underline{\boxed{\bf{a=0.44\:ms^{-2}}}}\:\orange{\bigstar}$

⚽ Distance travelled :

$\implies\sf\:v^2-u^2=2as\\ \\ \implies\sf\:(10)^2-(7.78)^2=2(0.44)s\\ \\ \implies\sf\:39.47=0.88s\\ \\ \implies\underline{\boxed{\bf{s=44.85m}}}\:\gray{\bigstar}$