Question

1 An athlete takes 2 second reach maximum
speed of 18 km/h from rest. What is the
magnitude of his average acceleration ?
(A) 1.5 m/s2
(B) 2.5 m/s2
(C) 3.5 m/s2
(D) 0.5 m/s2​

Answer 1

Given:-

→ Final velocity of the athlete = 18km/h

→ Time taken = 2s

To find:-

→ Magnitude of average acceleration

Solution:-

Firstly let's convert the final velocity of the athlete from km/h to m/s.

=> 1km/h = 5/18m/s

=> 18km/h = 5/18 × 18

=> 5m/s

In this case :-

• Initial velocity of the athlete will be zero

as the athlete was inititially at rest.

Now, let's calculalte the acceleration of the athlete using 1st equation of motion:-

=> v = u + at

Where:-

• v is final velocity of the athlete.

• u is initial velocity of the athlete.

• a is the acceleration of the athlete.

• t is time taken.

=> 5 = 0 + a(2)

=> 5 - 0 = 2a

=> 5 = 2a

=> a = 5/2

=> a = 2.5m/s²

Thus, average acceleration of the athlete is 2.5m/s² [ Option. (B) ].

Answer 2

Answer:

Given:-

$\sf {final\: velocity{(v}_{f})=18km/h }$

$\sf {initial\:velocity{(v}_{i})=0m/s}$

$\sf{Time taken=t =2s}$

To find:-

magnitude of average acceleration $\sf {A {}_{AVG}}$

Solution:-

• First convert v (f) from km/h to m/s

$\sf {{\cancel {18}}×{\dfrac {5}{{\cancel{18}}}}}$

$\longrightarrow$$\sf {5m/s}$

• Now as we know

$\sf {{\triangle}v=v {}_{f}-{v}_{i}}$

$\longrightarrow$$\sf {{\triangle}v=5-0}$

$\longrightarrow$$\sf {{\triangle}v=5}$

$\sf {{\triangle}t=2}$

• Now for magnitude we use formula

${\boxed{\sf {A{}_{AVG}={\dfrac {{\triangle{v}}}{{\triangle}t}}}}}$

• Substitute the values

$\sf {A {}_{AVG}={\dfrac {5}{2}}}$

$\longrightarrow$${\underline{\boxed{\bf {A {}_{AVG}=2.5m/s {}^{2}}}}}$

$\therefore$$\sf {Magnitude\:of\:average\:acceleration\:for\:} \\ \sf {the\:athlete\:is {\quad}{2.5m/s}^{2}.}$