Question

A car traveling initially at +7.0 m/s accelerates uniformly at the rate of +0.80 m/s² for a distance of 245 m.

Answer 1

Answer:

(a) 21 m/s

(b) 15.78 m/s

(c) 12.5 m/s

Explanation:

Given;

Initial velocity, u = +7.0 m/s

accelerates, a = +0.80 m/s²

distance, s = 245 m

One of equation of motions is given by;

                                $v^{2} = u^{2} + 2as$

where,

v is the final velocity in m/s

u is the initial velocity in m/s

a is the acceleration in m/s²

s is the distance in m

(a) The velocity of the car at end of the acceleration will be when it covers 245 m

                              $v^{2} = 7^{2} + (2 \ * 0.80s \ * 245)$

                              $v^{2} = 49 + 392$

                              $v^{2} = 441$

                              $v = \sqrt{441}$

                              $v = 21 \ m/s$

(b)  Velocity after the car accelerates for 125 m            

                            $v^{2} = 7^{2} + (2 \ * 0.80s \ * 125)$

                             $v^{2} = 49 + 200$

                             $v^{2} = 249$

                             $v = \sqrt{249}$

                             $v = 15.78 \ m/s$

(c)  Velocity after the car accelerates for 67 m                      

                             $v^{2} = 7^{2} + (2 \ * 0.80s \ * 67)$

                             $v^{2} = 49 + 107.2$

                             $v^{2} = 156.2$

                             $v = \sqrt{156.2}$

                             $v = 12.5 \ m/s$