Question

A cyclist is traveling at 7m s-1, then accelerates at 2.5m s-2 for 15 seconds.

a. How far did they travel during this time
b. What was the final speed after 15 seconds?

Answer 1

Answer:

(a) 386.25 m, (b) 44.5 m/s

Explanation:

u (initial velocity) = 7m/s

a (acceleration) = 2.5 m/s²

t (time) = 15 seconds

(a) Distance travelled (s)

s = ut + 1/2at²

⇒ (7*15) + (2.5*15*15)/2

⇒ 105 + 281.25

= 386.25 m

(b) Final velocity (v)

v = u + at

⇒ 7 + (2.5*15)

⇒ 7 + 37.5

= 44.5 m/s

Hence, the cyclist travelled 386.25 m far during this time, and his final speed was 44.5 m/s.

Answer 2

Given :-

• Initial Velocity (u) = 7 m/s
• Acceleration (a) = 2.5 m/s²
• Time = 15 seconds

To Find :-

• Distance (s) = ?
• Final Velocity (v) = ?

Solution :-

First of all we will find distance (s) travelled by the cyclist ,by using second equation of motion :

➻ s = ut + ½ at²

➻ s = 7 × 15 + ½ × 2.5 × (15)²

➻ s = 105 + ½ × 2.5 × 225

➻ s = 105 + ½ × 562.5

➻ s = 105 + 281.25

➻ s = 386.25 m

Now, let's find the Final Velocity of the cyclist by using first equation of motion :

➻ v = u + at

➻ v = 7 + 2.5 × 15

➻ v = 7 + 37.5

➻ v = 44.5 m/s

Hence,

• Distance travelled by cyclist is 386.25 m.
• and Final Velocity of the cyclist is 44.5 m/s.