Question

39. A motorcycle racer starts from rest and accelerates on a
straight track at 5m/sec. How far does the racer travel in 8
seconds?
(a) 100m
(b) 120m
(c) 140m
(d) 160m​

Answer 1

Given :

➨ Initial velocity = zero

➨ Acceleration = 5m/s²

➨ Time of journey = 8s

To Find :

➛ Distance covered by racer in the given interval of time.

Concept :

➜ Since, acceleration has said to be constant throughout the motion, we can easily apply second equation of kinematics to solve this question.

◈ Second equation of kinematics :

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

where,

✪ s denotes distance

✪ u denotes initial speed

✪ a denotes acceleration

✪ t denotes time

Calculation :

⇒ $\sf{s=(0\times 8)+\dfrac{1}{2}(5\times 8^2)}$

⇒ $\sf{s=0+\dfrac{1}{2}(320)}$

⇒ $\underline{\boxed{\bf{s=160m}}}$

Answer 2

$\huge\sf\pink{Answer}$

☞ Your Answer is Option D

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$\huge\sf\blue{Given}$

✭ Initial Velocity (u) = 0 m/s

✭ Acceleration (a) = 5 m/s²

✭ Time (t) = 8 seconds

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ The distance travelled?

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$\huge\sf\purple{Steps}$

Here we shall use the second Equation of motion,that is

$\underline{\boxed{\sf s = ut+ \dfrac{1}{2} at^2}}$

Substituting the given values,

➝ $\sf s = 0\times t +\dfrac{1}{2} \times 5 \times 8^2$

➝ $\sf s =0+ \dfrac{1}{2} \times 5 \times 64$

➝ $\sf s = 5 \times 32$

➝ $\sf \orange{s = 160 m}$

$\rule{170}3$