Question

A cyclist is travelling at 4.0 m/s. She speeds up to 16 m/s in a time of 5.6 s. Calculate her acceleration.​

Answer 1

Answer:

Initial velocity, u=15m/s

Initial velocity, u=15m/sFinal velocity, v=0m/s

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18m

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15)

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36a

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa=

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2 So, deceleration is 6.25m/s

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2 So, deceleration is 6.25m/s 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2 So, deceleration is 6.25m/s 2 .

Answer 2

Answer:

acceleration:  2.14 m/s²

Explanation:

          acceleration =  (final velocity - initial velocity ) ÷ time taken

Here the final velocity is 16 m/s, initial velocity is 4 m/s, time taken is 5.6 s

acceleration = ( 16 - 4 ) ÷ 5.6

                     → 12 ÷  5.6

                     → $\left[\begin{array}{ccc}\frac{15}{7}\end{array}\right]$ m/s² or 2.14 m/s²