Question

13. A bicycle moving with a velocity of 3 m/s
speeds up with an acceleration of 0.5 m/s2.
What will its velocity be after 5 s and how
far will it have moved during this time.
(Ans. 5.5 m/s, 21.25 m)​

Answer 1

u = 3m/s

a = 0.5m/s²

t = 5 s

By equation,

v = u+at

v = 3+0.5×5

v= 3+2.5

v = 5.5m/s

Again, By equation,

s = ut+½at²

s = 3×5+½×0.5×5²

s = 15+6.25

s = 21.25m

Cheers!

Answer 2

Answer:

The velocity of the bicycle and the distance moved by the bicycle after 5 seconds are 5.5m/s and 21.25m respectively.

Explanation:

For solving this question we will use the first and second equations of motion.

The values given in the question are:-

The initial velocity of the bicycle=3m/s

The acceleration of the bicycle=0.5m/s²

The time during which the motion occurs=5 seconds

CASE A: Final velocity

Using the first motion equation, we have,

$\mathrm{v=u+at}$       (1)

Where,

v=final velocity of the bicycle

u=initial velocity of the bicycle

a=acceleration of the bicycle

t=time during which the motion occurs

We obtain by entering the items in equation (1);

$\mathrm{v=3+0.5\times 5}$

$\mathrm{v=3+2.5}$

$\mathrm{v=5.5\ m/s}$    (2)

CASE B: Distance moved by bicycle

From the second equation of motion we have,

$\mathrm{S=ut+\frac{1}{2}at^2 }$        (3)

Where,

S=distance moved by bicycle

By inserting the values in equation (3) we get;

$\mathrm{S=3\times 5+\frac{1}{2}\times 0.5\times 5^2 }$

$\mathrm{S=15+6.25 }$

$\mathrm{S=21.25\ m}$     (4)

So, the velocity of the bicycle and the distance moved by the bicycle after 5 seconds are 5.5m/s and 21.25m respectively.

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