Question

A powerful motorcycle can accelerate from rest to 28 m/s in 4 seconds. What is its average acceleration? How far does it travel in that time?

Answer 1

Answer:

time t = 4 seconds

as it is starting from rest

initial velocity u = 0 m/s

final velocity v = 28m/s

Acceleration a= v-u/t

= 28-0/4

= 7m/s^2

distance s= ut+1/2at^2

s=0*4+1/2*7*(4)^2

= 1/2*7*16

= 7*8

=56 metres

Explanation:

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Answer 2

Given :

▪ Initial velocity = zero (i.e., rest)

▪ Final velocity = 28mps

▪ Time interval = 4s

To Find :

▪ Acceleration of motorcycle.

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

Concept :

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

↗ It is a vector quantity.

↗ It has both magnitude as well as direction.

↗ It can be positive, negative or zero.

↗ SI unit : m/s²

Calculation :

⚽ Acceleration :

$\dashrightarrow\bf\:a=\dfrac{v-u}{t}\\ \\ \dashrightarrow\sf\:a=\dfrac{28-0}{4}\\ \\ \dashrightarrow\underline{\boxed{\bf{\red{a=7\:ms^{-2}}}}}\:\gray{\bigstar}$

⚽ Distance travelled :

$\implies\bf\:v^2-u^2=2as\\ \\ \implies\sf\:(28)^2-(0)^2=2(7)s\\ \\ \implies\sf\:s=\dfrac{(28)^2}{14}\\ \\ \implies\sf\:s=2\times 28\\ \\ \implies\underline{\boxed{\bf{\green{s=56m}}}}\:\gray{\bigstar}$