Question

A man runs along the path shown in Figure 17. From point A to point B, he runs at a forward velocity of 4.5 m/s for 15.0 min. From point B to point C, he runs up a hill. He slows down at a constant rate of 0.050 m/s2 for 90.0 s and comes to a stop at point C. What was the total distance the man ran

Answer 1

The total distance ran by the man, d = 4252.5 m

Explanation:

Given data,

The velocity of man from A to B, v₁ = 4.5 m/s

The time period of his motion, t₁ = 15 min

                                                       = 900 s

B to C he slows down at a rate, a = -0.050 m/s²

The time period of his motion, t₂ = 90 s

The total distance covered by man A to C, d = ?

The distance from A to B,

                                 d₁ = v₁ x t₁

                                     = 4.5 m/s x 900 s

                                     = 4050 m

The distance from B to C

                             d₂ = |s| = ut + ½ at²

                                    = 4.5 x 90 + ½ (-0.050) 90²

                                     = 202.5 m

The total distance covered,

                               d = d₁ + d₂

                                  = 4050 m + 202.5 m

                                   = 4252.5 m

Hence, the total distance ran by the man, d = 4252.5 m

Answer 2

Total Distance travelled is 4252.5 m

Explanation:

Given data,

Man's velocity from A to B, $v_1 = 4.5\ m^-^1s$

The time period of his motion, $t_1$= 15 min

                                                           = $15 \times 60 s$

                                                           = 900 s

Retardation from B to C, a = $- 0.050\ ms^-^2$

The time period of motion for Man, $t_2$ = 90 s

The Total distance from A to C, d = ?

• The distance from A to B,

                               $d_1 = v_1 \times t_1$

                                    = $4.5\ ms^-^1 \times 900\ s$

                                    = 4050 m

• The distance  (B to C)

                 $d_2 = \left | s \right | = ut + \frac {1}{2} at^-^2$

                                   = $4.5 \times 90 + \frac {1}{2}(-0.050) 90^2$

                                    = 202.5 m

• The total distance ,

                              $d = d_1 + d_2$

                                 = 4050 m + 202.5 m

                                  = 4252.5 m

So, Total distance covered by man, d = 4252.5 m