A scooter moves with a speed of 7 ~ms ⁻¹ , on a straight road and is stopped by applying the brakes. Before stopping, the scooter travels 10 ~m . If the weight of the scooter is W , then the total resistance to the motion of the scooter will be

Step-by-step solution

Given initial speed u = 7 ~ms ⁻¹ , final speed v = 0 , and distance s = 10 ~m . Using the equation of motion v^2 = u^2 + 2as , we have 0 = 7^2 + 2a(10) . 0 = 49 + 20a a = - 49 20 ~ms ⁻² . The retarding force F is given by Newton's second law F = ma = W g a . Substituting the values, F = W 9.8 ( - 49 20 ) . Since 9.8 = 98 10 = 49 5 , we have F = W 49/5 ( - 49 20 ) = W 5 49 ( - 49 20 ) = - 5 20 W = - 1 4 W . The magnitude of the resistance force is 1 4 W . Answer: 1 4 W