An athlete, initially at rest, takes 2 s to reach his maximum speed of 36 km h- 1 . The magnitude of his average acceleration (in m s-2 ) is

Question

An athlete, initially at rest, takes 2 s to reach his maximum speed of 36 km h- 1 . The magnitude of his average acceleration (in m s-2 ) is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The final speed of the athlete, v=36 km h- 1=36× (5/18)=10 m s- 1 initial speed, u=0 time taken t=2 s for average acceleration, a v=u+at a=(v - u/t)=(10 - 0/2)=5 m s- 2

Q. An athlete, initially at rest, takes $2 s$ to reach his maximum speed of $36 km h^{- 1}$ . The magnitude of his average acceleration (in $m s^{- 2}$ ) is

The final speed of the athlete, $v = 36 km h^{- 1} = 36 \times \frac{5}{18} = 10 m s^{- 1}$
initial speed, $u = 0$
time taken $t = 2 s$
for average acceleration, $a$
$v = u + a t$
$a = \frac{v - u}{t} = \frac{10 - 0}{2} = 5 m s^{- 2}$

Q. An athlete, initially at rest, takes 2s to reach his maximum speed of 36kmh−1 . The magnitude of his average acceleration (in ms−2 ) is

The final speed of the athlete, v=36kmh−1=36×185​=10ms−1 initial speed, u=0 time taken t=2s for average acceleration, a v=u+at a=tv−u​=210−0​=5ms−2

Q. An athlete, initially at rest, takes $2 s$ to reach his maximum speed of $36 km h^{- 1}$ . The magnitude of his average acceleration (in $m s^{- 2}$ ) is

The final speed of the athlete, $v = 36 km h^{- 1} = 36 \times \frac{5}{18} = 10 m s^{- 1}$
initial speed, $u = 0$
time taken $t = 2 s$
for average acceleration, $a$
$v = u + a t$
$a = \frac{v - u}{t} = \frac{10 - 0}{2} = 5 m s^{- 2}$