Question

A truck covers 40.0 m in 9.45 s while uniformly slowing down to a final velocity of 2.35 m/s.

(a) Find the truck's original speed.
________m/s

(b) Find its acceleration.
_________ m/s²

A car starts from rest and travels for 7.5 s with a uniform acceleration of +3.0 m/s². The driver then applies the brakes, causing a uniform acceleration of −1.5 m/s². If the brakes are applied for 2.0 s, determine each of the following.

(c) How fast is the car going at the end of the braking period?
m/s

(d) How far has the car gone?
m

Answer 1

u a = +3.0 m/s2 a = -1.5m/sh initial velocity u= 0 u= us at = 3.0x 7.5 = 22.5 m/s is then acceleration a = – 1.5 mlr V2 = vt at = 22.5 - 1.58 2 = 19.5 m/s motion. @ in travelled distance first leg of - &= ut & l. at 2 2 0 + 2 x 3.0x 7.52 2 x = 84.375 m travelled in second leg. distance & 2 = 2. = 1,2 + 29.52 19.5² = 22.5² - 2x1.5xf2 $2= 42. m sa sits2 total distance - 84.4 - 42 126 m is 1

u a = +3.0 m/s2 a = -1.5m/sh initial velocity u= 0 u= us at = 3.0x 7.5 = 22.5 m/s is then acceleration a = – 1.5 mlr V2 = vt at = 22.5 - 1.58 2 = 19.5 m/s motion. @ in travelled distance first leg of - &= ut & l. at 2 2 0 + 2 x 3.0x 7.52 2 x = 84.375 m travelled in second leg. distance & 2 = 2. = 1,2 + 29.52 19.5² = 22.5² - 2x1.5xf2 $2= 42. m sa sits2 total distance - 84.4 - 42 126 m is 1

u²+baa = 2.352 t=9.45 solution :- &= 40 m, V=2.35 mls. final velocity 22= 42 + 2 as, 2.35² = 42 + 2.9x40 and ut at a x 9.45 2.35 = ut 4 + 9.45a = 2.35 solution for above equations 0= 14234 35721 mys, U = 23117 3780 [4= 6.11 mys, initial [a= - 0.4 mly2 acceleration.