Question

A motorcycle starts from rest and accelerates uniformly over a time of 4.89 seconds for a distance of 108 m. What is the acceleration of the motorcycle?

Answer 1

Answer:-

$\red{\bigstar}$$\large\boxed{\rm\purple{9.04 \: ms^{-2}}}$

• Given:-

Initial velocity [u] = 0 [as starts from rest]

Time [t] = 4.89 sec.

Distance covered [s] = 108m

• To Find:-

Acceleration [a] = ?

• Solution:-

We know,

• 2nd Equation of motion:-

$\pink{\bigstar}$$\large\boxed{\bf\blue{s = ut + \dfrac{1}{2} at^{2}}}$

Using the above equation of motion:-

here,

$\red{\dag}$ s = 108 m

$\red{\dag}$ u = 0

$\red{\dag}$ t = 4.89 seconds

$\dashrightarrow$ $\sf{108 = 0 \times 4.89 + \dfrac{1}{2} a \times (4.89)^{2}}$

$\dashrightarrow$$\sf{108 = 0 + \dfrac{1}{2} \times a \times 23.9121}$

$\dashrightarrow$$\sf{108 = a\times 11.95605}$

$\dashrightarrow$$\sf{a = \dfrac{108}{11.95605}}$

$\dashrightarrow$$\bf{a = 9.03308}$

$\dashrightarrow$$\bf\red{a = 9.04 \: ms^{-2}}$

Answer 2

Answer:

For first 10 seconds:

u = 0 m/s

f = 10 m/(s^2)

t = 10 s

Distance travelled in 10 seconds = u*t + (1/2)*f*(t^2) = 500 m.

Velocity after 10 seconds = u + f*t = 100 m/s.

For next 20 seconds:

Distance travelled in 20 seconds = (100*20) m = 2000 m.

Total distance travelled in 30 seconds = (500 + 2000) m = 2500 m.