Question

A cycle covered 2km in 8 minutes and initial velocity of the cycle was 1 m\ sec .find the acceleration that the cycle had in its motion​

Answer 1

Answer:

The acceleration of the cycle is 0.0132 m/s².

Explanation:

The distance covered by the cycle,S = 2 Km = 2000 m

The time taken to cover this distance, t = 8 min = 480 sec

The initial velocity of the cycle, u = 1 m/s

The acceleration can be calculated using the 2nd equation of motion.

                S = ut + (at² / 2)

∴ 2000 = (1 × 480) + [(a × 480²) / 2]

   2000 = 480 + (a × 115200)

2000 - 480 = a × 115200

1520 = a × 115200

   ∴ a = 1520 / 115200

      a = 0.0132 m/s²

Answer 2

Given: Distance covered by a cycle(S)=2km

            Time taken(t)=8 minutes

            Initial velocity(u)=1m/sec

To find The acceleration that the cycle had in its motion​.

Solution: A body in which there is a change in the rate of velocity change from the initial point to the final point with respect to the time taken is commonly termed as the body in accelerated motion. The S.I. unit of this physical quantity is metre per second square(m/sec²). It is actually a vector quantity that has both direction and magnitude.

From above S=2km=(2×1000)m [∵ 1km=1000m]

                              =2000m  

t=8 minutes=(8×60)sec  [∵1min =60sec]

                  =480sec

u=1m/sec

We know that for the motion in one dimensional, we have

  S=ut+(1/2)at²  [a= acceleration]

⇒2000=(1×480)+(1/2)×a×(480)² [substituting the values of S,t, and u]

⇒2000-480=(1×a×240×480)  [∵480/2=240]

⇒1520=115200a

⇒115200a=1520  [rearranging the sides]

⇒a=1520/115200

⇒a=0.01319m/sec²

     ≅0.0132m/sec²

Hence the acceleration that the cycle had in its motion​ is 0.0132m/sec².