A truck running at 90 text{ km h}^{-1} is bro | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: Initial velocity $u = 90 	ext{ km h}^{-1} = 90 	imes \frac{5}{18} 	ext{ m s}^{-1} = 25 	ext{ m s}^{-1}$.Final velocity $v = 0 	ext{ m

Vedclass · Accepted Answer

Retardation = $12.5 	ext{ m s}^{-2}$,Time = $2 	ext{ s}$

Vedclass · Answer

(A) Given: Initial velocity $u = 90 	ext{ km h}^{-1} = 90 	imes \frac{5}{18} 	ext{ m s}^{-1} = 25 	ext{ m s}^{-1}$.Final velocity $v = 0 	ext{ m s}^{-1}$ (as the truck comes to rest).Distance $S = 25 	ext{ m}$.$(i)$ To find retardation $(a)$: Using the third equation of motion,$v^{2} - u^{2} = 2aS$.$(0)^{2} - (25)^{2} = 2 	imes a 	imes 25$.$-625 = 50a$.$a = -\frac{625}{50} = -12.5 	ext{ m s}^{-2}$.Retardation is the magnitude of negative acceleration,so retardation = $12.5 	ext{ m s}^{-2}$.(ii) To find time $(t)$: Using the first equation of motion,$v = u + at$.$0 = 25 + (-12.5) 	imes t$.$12.5t = 25$.$t = \frac{25}{12.5} = 2 	ext{ s}$.

$A$ truck running at $90 \text{ km h}^{-1}$ is brought to rest over a distance of $25 \text{ m}$. Calculate the retardation and time for which brakes are applied.

Similar Questions

$A$ racing car has a uniform acceleration of $4 \,m s^{-2}$. What distance (in $m$) will it cover in $10 \,s$ after start?

$A$ body moving with uniform acceleration has velocities $20 \, m/s$ and $30 \, m/s$ when passing two points $A$ and $B$. Then the velocity midway between $A$ and $B$ is:

$A$ ball is gently dropped from a height of $20 \, m$. If its velocity increases uniformly at the rate of $10 \, m s^{-2}$,with what velocity will it strike the ground? After what time will it strike the ground?

$A$ scooter starts from rest and moves in a straight line with a constant acceleration,covering a distance of $64 \, m$ in $4 \, s$.
$(i)$ Calculate its acceleration and its final velocity.
$(ii)$ At what time had the scooter covered half the total distance?

Two stones are thrown vertically upwards simultaneously with their initial velocities $u_{1}$ and $u_{2}$ respectively. Prove that the heights reached by them would be in the ratio of $u_{1}^{2}: u_{2}^{2}$ (Assume upward acceleration is $-g$ and downward acceleration is $+g$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module